Integration methods for rational fractions with square roots

Integrals containing \(\sqrt{a^2 - t^2}\)

1. Fraction with a root at the denominator\(: \frac{1}{\sqrt{a^2 - t^2}}\)

   $$ \forall a \in \hspace{0.04em} \mathbb{R}, \ \forall x \in \hspace{0.04em} \Bigl]-|a|, \hspace{0.2em} |a| \Bigr[, \ I_1(x) = \frac{1}{\sqrt{a^2 - x^2}} $$
   $$ \int^x I_1(t) \ dt = \int^x \frac{dt}{\sqrt{a^2 - t^2}} $$
   $$ \int^x I_1(t) \ dt = \int^x \frac{ |a| \ \cos(u) \ du }{\sqrt{a^2 - |a|^2 \sin^2(u)}}$$

   $$\text{with } \Biggl \{ \begin{gather*} t = |a| \ \sin(u) \\ dt = |a| \ \cos(u) \ du \end{gather*}$$
   $$ \int^x I_1(t) \ dt = \frac{|a|}{|a|} \int^x \frac{ \cos(u) \ du}{\sqrt{1 - \sin^2(u)}} $$
   $$ \int^x I_1(t) \ dt = \int^x \frac{\cos(u)}{\sqrt{\cos^2(u)}} \ du $$
   $$ \int^x I_1(t) \ dt = \int^x \frac{\cos(u)}{\left|\cos(u)\right|} \ du \qquad \left(\text{with } u \neq \frac{\pi}{2} \Longleftrightarrow t \neq 0, \text{ which is the case} \right) $$

   We have a sign generator placed in the integrand, let's study its value in the study interval \(\hspace{0.04em} ]-a, \hspace{0.2em} a[\).

   However, we know from the derivative of a reciprocal function that:

   $$ \forall (f,f^{-1}), \enspace (f' \circ f^{-1}) \neq 0, $$

   $$ ( f^{-1} )' = \frac{1}{ (f' \circ f^{-1})} $$

   So in our case,

   $$ \left( \operatorname{Arcsin}(x) \right)' = \frac{1}{ \cos\left(\operatorname{Arcsin}(x)\right)} $$

   $$\text{with } \Biggl \{ \begin{gather*} f^{-1} = \operatorname{Arcsin}(x) \\ f = \sin(x) \Longrightarrow f' = \cos(x) \end{gather*}$$

   $$ \Longleftrightarrow \ \cos\left(\operatorname{Arcsin}(x)\right) = \frac{1}{ \left( \operatorname{Arcsin}(x) \right)' } $$

   But we know the derivative of the \(\operatorname{Arcsin}(x)\) function :

   $$ \forall x \in \hspace{0.04em} ]-1 ,\hspace{0.2em} 1[, $$

   $$ \operatorname{Arcsin}(x)' = \frac{1}{\sqrt{1 - x^2}} $$

   So, by combining the two previous expressions we obtain:

   $$ \cos\left(\operatorname{Arcsin}(x)\right) = \sqrt{1 - x^2} \hspace{4em}\Big[ \cos(\operatorname{Arcsin})\Bigr] $$

   By replacing our initial variable in this sign generator, we have:

   $$ \frac{\cos(u)}{\left|\cos(u)\right|} = \frac{\sqrt{1 - \left( \frac{x}{|a|}\right)^2}}{\left|\sqrt{1 - \left( \frac{x}{|a|}\right)^2}\right| }$$
   $$ \frac{\cos(u)}{\left|\cos(u)\right|} = \frac{\sqrt{ \frac{a^2 - x^2}{a^2}}}{\left|\sqrt{ \frac{a^2 - x^2}{a^2}}\right| }$$
   $$ \frac{\cos(u)}{\left|\cos(u)\right|} = \frac{\sqrt{a^2 - x^2}}{\left|\sqrt{a^2 - x^2}\right| }$$

   In the study interval, this ratio is always worth \(1\).

   Now we integrate only:

   $$ \int^x I_1(t) \ dt = \int^x du $$
   $$ \int^x I_1(t) \ dt = \Bigl[ u \Bigr]^{u = \operatorname{Arcsin}\left( \frac{x}{|a|}\right)} $$

   So,

   $$ \int^x I_1(t) \ dt = \operatorname{Arcsin}\left( \frac{x}{|a|}\right)$$

   And finally,

   $$ \forall a \in \hspace{0.04em} \mathbb{R}^*, \ \forall x \in \hspace{0.04em} \Bigl [-|a|, \hspace{0.2em} |a| \Bigr], $$
   $$ \int^x \frac{dt}{\sqrt{a^2 - t^2}} = \operatorname{Arcsin}\left(\frac{x}{|a|}\right) $$

2. Fraction with \(t\) and a root in the denominator\(: \frac{1}{t\sqrt{a^2 - t^2}}\)

   $$ \forall a \in \hspace{0.04em} \mathbb{R}, \ \forall x \in \hspace{0.04em} \Bigl]-|a|, \hspace{0.2em} 0 \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]0, \hspace{0.2em} |a| \Bigr[ , \ I_2(x) = \frac{1}{x\sqrt{a^2 - x^2}} $$
   $$ \int^x I_2(t) \ dt = \int^x \frac{dt}{t\sqrt{a^2 - t^2}} $$

   We set down a new variable: \(u = \sqrt{a^2 - t^2}\).

   $$ \int^x I_2(t) \ dt = -\int^x \frac{du}{t^2} $$

   $$\text{with } \Biggl \{ \begin{gather*} u = \sqrt{a^2 - t^2} \\ du = -\frac{t}{\sqrt{a^2 - t^2}} \ dt \end{gather*}$$

   But,

   $$u = \sqrt{a^2 - t^2} \Longleftrightarrow t^2 = a^2 - u^2$$

   So,

   $$ \int^x I_2(t) \ dt = -\int^x \frac{du}{a^2 - u^2} = \int^x \frac{du}{u^2 - a^2} $$

   Let us decompose this integrand in simple elements .

   $$F(u) = \frac{1}{u^2 - a^2}$$

   $$F(u) = \frac{1}{(u-a)(u+a)}$$

   Then, it exists \((\alpha, \beta) \in \hspace{0.04em}\mathbb{R}^2\) such as:

   $$F(u) = \frac{\alpha}{u-a} + \frac{\beta}{u+a}$$

   By performing \( (u = a)\), we determine \( \alpha \):

   $$ \underset{(u=a)}{F(u)} (u-a) = \frac{1}{(u+a)} = \alpha \Longrightarrow \left( \alpha = \frac{1}{2a} \right) $$

   Now by performing \( (u = -a)\), we determine \( \beta \):

   $$ \underset{(u=-a)}{F(u)} (u+a) = \frac{1}{(u-a)} = \beta \Longrightarrow \left( \beta = -\frac{1}{2a} \right) $$

   So, the the integral can now be rewritten as:

   $$ \int^x I_2(t) \ dt = \frac{1}{2a} \int^x\frac{1}{u-a}\ du - \frac{1}{2a} \int^x\frac{1}{u+a} \ du $$
   $$ \int^x I_2(t) \ dt = \frac{1}{2a} \Bigl[ \ln|u-a|\Bigr]^{u = \sqrt{a^2 - x^2}} - \frac{1}{2a} \Bigl[ \ln|u+a|\Bigr]^{u = \sqrt{a^2 - x^2}} \qquad(I_2)$$
   $$ \int^x I_2(t) \ dt = \frac{1}{2a} \ln\left|\sqrt{a^2 - x^2} - a \right| + \frac{1}{2a} \ln\left|\sqrt{a^2 -x^2} + a\right| $$

   And as a result,

   $$ \forall a \in \hspace{0.04em} \mathbb{R}, \ \forall x \in \hspace{0.04em} \Bigl]-|a|, \hspace{0.2em} 0 \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]0, \hspace{0.2em} |a| \Bigr[ , $$
   $$ \int^x \frac{dt}{t\sqrt{a^2 - t^2}} = \frac{1}{2a} \ln\left|\sqrt{a^2 - x^2} - a \right| + \frac{1}{2a} \ln\left|\sqrt{a^2 - x^2} + a\right| $$

3. Fraction with \(t^2\) and a root in the denominator\(: \frac{1}{t^2\sqrt{a^2 - t^2}}\)

   $$ \forall a \in \hspace{0.04em} \mathbb{R}, \ \forall x \in \hspace{0.04em} \Bigl]-|a|, \hspace{0.2em} 0 \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]0, \hspace{0.2em} |a| \Bigr[ , \ I_3(x) = \frac{1}{x^2\sqrt{a^2 - x^2}} $$
   $$ \int^x I_3(t) \ dt = \int^x \frac{dt}{t^2\sqrt{a^2 - t^2}} $$

   We set down : \(t = |a| \ \sin(u)\).

   $$ \int^x I_3(t) \ dt = \int^x \frac{|a| \ \cos(u) \ du }{ |a|^2 \ \sin^2(u)\sqrt{a^2 - |a|^2 \ \sin^2(u)}} $$

   $$\text{with } \Biggl \{ \begin{gather*} t = |a| \ \sin(u) \\ dt = |a| \ \cos(u) \ du \end{gather*}$$

   Then we have,

   $$ \int^x I_3(t) \ dt = \frac{1}{a^2} \int^x \frac{\ \cos(u) \ du}{\sin^2(u) \sqrt{\cos^2(u)} } $$
   $$ \int^x I_3(t) \ dt = \frac{1}{a^2} \int^x \frac{\ \cos(u) \ du}{\sin^2(u) \left|\cos(u)\right| } $$
   $$ \int^x I_3(t) \ dt = \frac{1}{a^2} \int^x \frac{\ \cos(u) }{\left|\cos(u)\right| } \ \csc^2(u) \ du $$

   As well as above, this sign generator always worth \(1\):

   $$ \forall x \in \hspace{0.04em} ]-a, \hspace{0.2em} 0[ \hspace{0.04em} \cup \hspace{0.04em} ]0, \hspace{0.2em} a[ , \frac{\ \cos(u) }{\left|\cos(u)\right| } = 1 $$

   Therefore, we only have to integrate:

   $$ \int^x I_3(t) \ dt = \frac{1}{a^2} \int^x \csc^2(u) \ du $$

   We are in the case of a standard antiderivative .

   $$ \int^x \csc^2(t) \ dt = -\cot(x) $$

   We replace it by its value.

   $$ \int^x I_3(t) \ dt = - \frac{1}{a^2} \cot(u) $$
   $$ \int^x I_3(t) \ dt = - \frac{1}{a^2} \cot\left(\operatorname{Arcsin}\left( \frac{x}{|a|}\right)\right) $$
   $$ \int^x I_3(t) \ dt = - \frac{1}{a^2} \frac{\cos\left(\operatorname{Arcsin}\left( \frac{x}{|a|}\right)\right)}{\sin\left(\operatorname{Arcsin}\left( \frac{x}{|a|}\right)\right)} $$
   $$ \int^x I_3(t) \ dt = - \frac{1}{a^2} \frac{\cos\left(\operatorname{Arcsin}\left( \frac{x}{|a|}\right)\right)}{\left( \frac{x}{|a|}\right)} $$

   We use again \(\Bigl[ \cos(\operatorname{Arcsin})\Bigr] \):

   $$ \cos\left(\operatorname{Arcsin}(x)\right) = \sqrt{1 - x^2} \hspace{4em}\Big[ \cos(\operatorname{Arcsin})\Bigr] $$
   $$ \int^x I_3(t) \ dt = - \frac{|a|}{|a|^2} \frac{1}{x} \sqrt{1 - \left( \frac{x}{|a|}\right)^2}$$
   $$ \int^x I_3(t) \ dt = - \frac{1}{a^2} \frac{\sqrt{a^2 - x^2}}{x} $$

   And finally we do obtain,

   $$ \forall a \in \hspace{0.04em} \mathbb{R}, \ \forall x \in \hspace{0.04em} \Bigl]-|a|, \hspace{0.2em} 0 \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]0, \hspace{0.2em} |a| \Bigr[ ,$$
   $$ \int^x \frac{dt}{t^2\sqrt{a^2 - t^2}} = - \frac{1}{a^2} \frac{\sqrt{a^2 - x^2}}{x} $$

4. Simple root\(: \sqrt{a^2 - t^2} \)

   $$ \forall a \in \hspace{0.04em} \mathbb{R}, \forall x \in \hspace{0.04em} \Bigl[-|a|, \hspace{0.2em} |a| \Bigr], \ I_4(x) = \sqrt{a^2 - x^2} $$
   $$ \int^x I_4(t) \ dt = \int^x \sqrt{a^2 - t^2} \ dt $$
   $$ \int^x I_4(t) \ dt = \int^x \sqrt{a^2 - |a|^2 \sin^2(u)} \ |a| \ \cos(u) \ du $$

   $$\text{with } \Biggl \{ \begin{gather*} t = |a| \ \sin(u) \\ dt = |a| \ \cos(u) \ du \end{gather*}$$
   $$ \int^x I_4(t) \ dt = a^2 \int^x \sqrt{\cos^2(u)} \ \cos(u) \ du$$
   $$ \int^x I_4(t) \ dt = a^2 \int^x |\cos(u) | \ \cos(u) \ du$$
   $$ \int^x I_4(t) \ dt = a^2 \int^x \frac{\cos(u) \ \cos^2(u)}{|\cos(u) |} \ du$$

   As well as above, this sign generator always worth \(1\):

   $$ \forall x \in \hspace{0.04em} ]-a, \hspace{0.2em} 0[ \hspace{0.04em} \cup \hspace{0.04em} ]0, \hspace{0.2em} a[ , \frac{\ \cos(u) }{\left|\cos(u)\right| } = 1 $$

   And now we integrate this:

   $$ \int^x I_4(t) \ dt = a^2 \int^x \cos^2(u) \ du$$

   To integrate this trig power, we will use the trigonometric duplication formulas to linearize it.

   $$\forall x \in \mathbb{R},$$

   $$ \cos(2\alpha) = 2 \cos^2(\alpha) -1 \ \Longleftrightarrow \ \cos^2(\alpha) = \frac{1 + \cos(2\alpha)}{2} $$

   So,

   $$ \int^x I_4(t) \ dt = a^2 \int^x \frac{1 + \cos(2u)}{2} \ du$$
   $$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl[ \int^x du + \int^x \cos(2u) \ du \Biggr]$$
   $$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl[ u + \frac{\sin(2u)}{2} \Biggr]^u$$

   We use another trigonometric duplication formula to linearize it.

   $$\forall x \in \mathbb{R},$$

   $$ \sin(2\alpha) = 2sin(\alpha)\cos(\alpha) $$
   $$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl[ u + \frac{2sin(u)\cos(u)}{2} \Biggr]^u$$
   $$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl[ u + \sin(u)\cos(u) \Biggr]^{u = \operatorname{Arcsin}\left( \frac{x}{|a|} \right)} $$

   As previously, replacing \(u\) by its value:

   $$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl( \operatorname{Arcsin}\left(\frac{x}{|a|}\right) + \frac{x}{|a|} \ \cos \left(\operatorname{Arcsin}\left(\frac{x}{|a|}\right)\right) \Biggr) $$

   But, we already seen several times that:

   $$ \cos\left(\operatorname{Arcsin}(x)\right) = \sqrt{1 - x^2} \hspace{4em}\Big[ \cos(\operatorname{Arcsin})\Bigr]$$

   By injecting it, we have now:

   $$ \int^x I_4(t) \ dt = \frac{a^2}{2} \Biggl( \operatorname{Arcsin}\left(\frac{x}{|a|}\right) + \frac{x}{|a|}\sqrt{1 - \left(\frac{x}{|a|}\right)^2} \Biggr)$$
   $$ \int^x I_4(t) \ dt = \frac{a^2}{2} \operatorname{Arcsin}\left(\frac{x}{|a|}\right) + \frac{a^2x}{2|a|^2} \sqrt{a^2 - x^2} $$

   And as a result,

   $$ \forall a \in \hspace{0.04em} \mathbb{R}^*, \ \forall x \in \hspace{0.04em} \Bigl[-|a|, \hspace{0.2em} |a| \Bigr], $$
   $$ \int^x \sqrt{a^2 - t^2} \ dt = \frac{a^2}{2} \operatorname{Arcsin}\left(\frac{x}{|a|}\right) + \frac{x}{2} \sqrt{a^2 - x^2} $$

Integrals containing \(\sqrt{a^2 + t^2}\)

1. Fraction with a root at the denominator\(: \frac{1}{\sqrt{a^2 + t^2}}\)

   $$ \Big[ \forall (a,x) \in \hspace{0.04em} \big[\mathbb{R}^* \bigr]^2 \Bigr] \lor \Big[ (a = 0) \land (x \neq 0) \Bigr], \ J_1(x) = \frac{1}{\sqrt{a^2 + x^2}} $$
   $$ \int^x J_1(t) \ dt = \int^x \frac{dt}{\sqrt{a^2 + t^2}} $$

   1. Setting down \( t = |a| \ \sinh(u)\)

      By setting down \( t = |a| \ \sinh(u)\), we do have:

      $$ \int^x J_1(t) \ dt = \int^x \frac{|a| \ \cosh(u) \ du }{\sqrt{a^2 + |a|^2sinh^2(u)}} $$

      $$\text{with } \Biggl \{ \begin{gather*} t = |a| \ \sinh(u) \\ dt = |a| \ \cosh(u) \ du \end{gather*}$$
      $$ \int^x J_1(t) \ dt = \frac{|a|}{|a|} \int^x \frac{\cosh(u) \ du}{\sqrt{\cosh^2(u)}} $$
      $$ \int^x J_1(t) \ dt = \int^x \frac{\cosh(u) \ du}{|\cosh(u)|} $$

      We again have a sign generator placed in the integrand, let's study its value in the study interval \( \mathbb{R}^*\).

      The same way as:

      $$ \cos\left(\operatorname{Arcsin}(x)\right) = \sqrt{1 - x^2} \hspace{4em}\Big[ \cos(\operatorname{Arcsin})\Bigr]$$

      The equivalent for hyperbolic functions is:

      $$ \cosh\left(\operatorname{Argsinh}(x)\right) = \sqrt{1 + x^2} \hspace{4em}\Big[ \cosh(\operatorname{Argsinh})\Bigr]$$

      By replacing our initial variable in this sign generator, we have:

      $$ \frac{\cosh(u)}{\left|\cosh(u)\right|} = \frac{\sqrt{1 + \left( \frac{x}{|a|}\right)^2}}{\left|\sqrt{1 + \left( \frac{x}{|a|}\right)^2}\right| }$$
      $$ \frac{\cosh(u)}{\left|\cosh(u)\right|} = \frac{\sqrt{ \frac{a^2 + x^2}{a^2}}}{\left|\sqrt{ \frac{a^2 + x^2}{a^2}}\right| }$$
      $$ \frac{\cosh(u)}{\left|\cosh(u)\right|} = \frac{\sqrt{a^2 + x^2}}{\left|\sqrt{a^2 + x^2}\right| }$$

      In any case, this ration is always worth \(1\).

      So, we now integrate:

      $$ \int^x J_1(t) \ dt = \int^x du $$
      $$ \int^x J_1(t) \ dt = \Bigl[ u \Bigr]^{u = \operatorname{Argsinh}\left( \frac{x}{|a|}\right)} $$

      And as a result,

      $$ \forall (a, x) \in \hspace{0.04em} \mathbb{R}^2, \ (a \neq 0 \lor x \neq 0), $$

      $$ \int^x \frac{dt}{\sqrt{a^2 + t^2}} = \operatorname{Argsinh}\left(\frac{x}{|a|} \right) $$

      $$(4)$$

   2. Setting down \( t = |a| \ \tan(u)\)

      Moreover, by setting down \( t = |a| \ \tan(u)\):

      $$ \int^x J_1(t) \ dt = \int^x \frac{ |a| \ \sec^2(u) \ du }{\sqrt{a^2 + |a|^2tan^2(u)}} $$

      $$\text{with } \Biggl \{ \begin{gather*} t = |a| \ \tan(u) \\ dt = |a| \ (1 + \tan^2(u)) \ du = |a| \ \sec^2(u) \ du \end{gather*}$$
      $$ \int^x J_1(t) \ dt = \frac{|a|}{|a|} \int^x \frac{\sec^2(u) \ du }{ \sqrt{\sec^2(u)}} $$
      $$ \int^x J_1(t) \ dt = \int^x \frac{\sec^2(u) \ du }{ |\sec(u)|} $$

      We again have a sign generator placed in the integrand, let's study its value in the study interval \( \mathbb{R}^*\).

      Using again the derivative of a reciprocal function , we do have:

      $$ \left( \operatorname{Arctan}(x) \right)' = \frac{1}{ \sec^2\left(\operatorname{Arctan}(x)\right)} $$

      $$\text{with } \Biggl \{ \begin{gather*} f^{-1} = \operatorname{Arctan}(x) \\ f = \tan(x) \Longrightarrow f' = \sec^2(x) \end{gather*}$$

      $$ \Longleftrightarrow \ \sec^2(\operatorname{Arctan}(x)) = \frac{1}{ \left( \operatorname{Arctan}(x) \right)' } $$

      But we know the derivative of the \(\operatorname{Arctan}(x)\) function :

      $$ \forall x \in \mathbb{R}, $$

      $$ \operatorname{Arctan}(x)' = \frac{1}{1 + x^2} $$

      By combining the two previous expressions, we obtain:

      $$\sec^2(\operatorname{Arctan}(x)) = 1+ x^2 \Longleftrightarrow \sec\left(\operatorname{Arctan}(x)\right) = \sqrt{ 1 + x^2 } \hspace{4em} \Big[\sec^2(\operatorname{Arctan}) \Bigr]$$

      By replacing our initial variable in this sign generator, we have:

      $$ \frac{\sec(u)}{\left|\sec(u)\right|} = \frac{\sqrt{1 + \left( \frac{x}{|a|}\right)^2}}{\left|\sqrt{1 + \left( \frac{x}{|a|}\right)^2}\right| } = \frac{\sqrt{a^2 + x^2}}{\left|\sqrt{a^2 + x^2}\right| }$$

      In any case, this ration is always worth \(1\).

      So, we only integrate:

      $$ \int^x J_1(t) \ dt = \int^x \sec(u) \ du $$

      However, we do have below in the page, in the trigonometric antiderivatives that:

      $$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \Bigl \{ \frac{\pi}{2} + k\pi \Bigr \} \biggr], $$

      $$\int^x \sec(t) \ dt = \ln \left|\sec(x) + \tan(x) \right|$$
      $$ \int^x J_1(t) \ dt = \Bigl [ \ln \left|\sec(u) + \tan(u) \right| \Bigr]^{u = \operatorname{Arctan}\left( \frac{x}{|a|} \right)} $$
      $$ \int^x J_1(t) \ dt = \ln \left|\sec\left(\operatorname{Arctan}\left( \frac{x}{|a|} \right)\right) + \tan\left(\operatorname{Arctan}\left( \frac{x}{|a|} \right)\right) \right|$$

      Using again the derivative of a reciprocal function , we do have:

      $$ \left( \operatorname{Arctan}(x) \right)' = \frac{1}{ \sec^2\left(\operatorname{Arctan}(x)\right)} $$

      $$\text{with } \Biggl \{ \begin{gather*} f^{-1} = \operatorname{Arctan}(x) \\ f = \tan(x) \Longrightarrow f' = \sec^2(x) \end{gather*}$$

      $$ \Longleftrightarrow \ \sec^2(\operatorname{Arctan}(x)) = \frac{1}{ \left( \operatorname{Arctan}(x) \right)' } $$

      But we know the derivative of the \(\operatorname{Arctan}(x)\) function :

      $$ \forall x \in \mathbb{R}, $$

      $$ \operatorname{Arctan}(x)' = \frac{1}{1 + x^2} $$

      By combining the two previous expressions, we obtain:

      $$\sec^2(\operatorname{Arctan}(x)) = 1+ x^2 \Longleftrightarrow \sec\left(\operatorname{Arctan}(x)\right) = \sqrt{ 1 + x^2 } \hspace{4em} \Big[\sec^2(\operatorname{Arctan}) \Bigr]$$

      By then injecting our result into the initial expression we have:

      $$ \int^x J_1(t) \ dt = \ln \left|\sqrt{ 1 + \left(\frac{x}{|a|}\right)^2 } + \frac{x}{|a|} \right| $$
      $$ \int^x J_1(t) \ dt = \ln \left|\frac{1}{|a|}\sqrt{ a^2 + x^2 } + \frac{x}{|a|} \right| $$
      $$ \int^x J_1(t) \ dt = \ln \left|\sqrt{ a^2 + x^2 } + x\right| - \ln \Bigl| |a| \Bigr| $$

      The constant \(- \ln \Bigl| |a| \Bigr| \) being absorbed by the main integration constant, we finally obtain that:

      $$ \Bigl[ \forall (a,x) \in \hspace{0.04em} \mathbb{R}^2 \Bigr] \hspace{0.2em} \backslash \hspace{0.2em} \Bigl \{ (0,0) \Bigr \} , $$

      $$ \int^x \frac{dt}{\sqrt{a^2 + t^2}} = \ln \left|\sqrt{ a^2 + x^2 } + x\right|$$

      $$(5)$$

   Both expressions \((4)\) and \((5)\) having a common member, they are equal up to a constant.

   $$ \operatorname{Argsinh}\left( \frac{x}{|a|}\right) + C_1 = \ln \left| \sqrt{ x^2 + a^2 } + x \right| + C_2 $$

   Let us determine this constant by taking a value of \(x\), for example \(x = 0\).

   $$ \operatorname{Argsinh}\left( \frac{0}{|a|}\right) = 0 $$

   $$\ln \left| \sqrt{0^2+ \ a^2 } + 0 \right| = \ln(a)$$

   So, we find that:

   $$ (C_1 = \ln|a| + C_2 ) \Longrightarrow (C_1 - C_2 = \ln|a|)$$
   $$ \operatorname{Argsinh}\left( \frac{x}{|a|}\right) + \ln|a| = \ln \left| \sqrt{ x^2 - a^2 } + x \right| $$

   We then have as a bonus an explicit definition of the \(acrsinh\) function with \(a = 1\):

   $$ \forall x \in \mathbb{R}, $$
   $$ \operatorname{Argsinh}(x) = \ln \left| x + \sqrt{ 1 + x^2 } \right| $$

2. Fraction with \(t\) and a root in the denominator\(: \frac{1}{t\sqrt{a^2 + t^2}}\)

   $$ \forall a \in \mathbb{R}, \ \forall x \in \hspace{0.04em} \mathbb{R}^*, \ J_2(x) = \frac{1}{x\sqrt{a^2 + x^2}} $$
   $$ \int^x J_2(t) \ dt = \int^x \frac{dt}{t\sqrt{a^2 + t^2}} $$

   As previously, we set down: \(u = \sqrt{a^2 + t^2}\).

   $$ \int^x J_2(t) \ dt = \int^x \frac{du}{t^2} $$

   $$\text{with } \Biggl \{ \begin{gather*} u = \sqrt{a^2 + t^2} \\ du = \frac{t}{\sqrt{a^2 + t^2}} \ dt \end{gather*}$$

   Or,

   $$u = \sqrt{a^2 + t^2} \Longleftrightarrow t^2 = u^2 - a^2$$

   Now we integrate the following:

   $$ \int^x J_2(t) \ dt = \int^x \frac{du}{u^2 - a^2} $$

   But, we already solved this integral above:

   $$ \int^x \frac{du}{u^2 - a^2} = \frac{1}{2a} \Bigl[ \ln|u-a|\Bigr]^{u = \sqrt{a^2 - x^2}} - \frac{1}{2a} \Bigl[ \ln|u+a|\Bigr]^{u = \sqrt{a^2 - x^2}} \qquad(I_2) $$
   $$ \int^x J_2(t) \ dt = \frac{1}{2a} \Bigl[ \ln|u-a|\Bigr]^{u = \sqrt{a^2 + x^2}} - \frac{1}{2a} \Bigl[ \ln|u+a|\Bigr]^{u = \sqrt{a^2 + x^2}} $$
   $$ \int^x J_2(t) \ dt = \frac{1}{2a} \ln\left|\sqrt{a^2 + x^2}-a\right| - \frac{1}{2a} \ln \left|\sqrt{a^2 + x^2} + a\right| $$

   And finally,

   $$ \forall a \in \hspace{0.04em} \mathbb{R}^+, \ \forall x \in \hspace{0.04em} \mathbb{R}^*,$$
   $$ \int^x \frac{dt}{t\sqrt{a^2 + t^2}} = \frac{1}{2a} \ln\left|\sqrt{a^2 + x^2}-a\right| - \frac{1}{2a} \ln \left|\sqrt{a^2 + x^2} + a\right| $$

3. Fraction with \(t^2\) and a root in the denominator\(: \frac{1}{t^2\sqrt{a^2 + t^2}}\)

   $$ \forall a \in \mathbb{R}, \ \forall x \in \hspace{0.04em} \mathbb{R}^*, \ J_3(x) = \frac{1}{x^2\sqrt{a^2 + x^2}} $$
   $$ \int^x J_3(t) \ dt = \int^x \frac{dt}{t^2\sqrt{a^2 + t^2}} $$

   We set down the new variable: \(t = |a| \ \tan(u)\).

   $$ \int^x J_3(t) \ dt = \int^x \frac{|a| \ \sec^2(u) \ du }{ |a|^2 \ \tan^2(u)\sqrt{a^2 + |a|^2 \ \tan^2(u)}} $$

   $$\text{with } \Biggl \{ \begin{gather*} t = |a| \ \tan(u) \\ dt = |a| \ \left(1 + \tan^2(u)\right) \ du = |a| \ \sec^2(u) \ du \end{gather*}$$

   Which gives us,

   $$ \int^x J_3(t) \ dt = \frac{1}{a^2} \int^x \frac{\sec^2(u) }{\tan^2(u)\sqrt{\sec^2(u)}} \ du $$
   $$ \int^x J_3(t) \ dt = \frac{1}{a^2} \int^x \frac{\sec(u)}{|\sec(u)|} \frac{\sec(u)}{\tan^2(u)} \ du $$

   We saw above that this sign generator was always worth \(1\):

   $$ \forall x \in \hspace{0.04em} \mathbb{R}^*, \ \frac{\sec(u)}{|\sec(u)|} = 1 $$

   Consequently, removing it from the integrand:

   $$ \int^x J_3(t) \ dt = \frac{1}{a^2} \int^x \frac{\sec(u)}{\tan^2(u)} \ du $$
   $$ \int^x J_3(t) \ dt = \frac{1}{a^2} \int^x \frac{\cos(u)}{\sin^2(u)} \ du $$

   We set another variable down: \(v =\sin(u)\). We do have:

   $$ \int^x J_3(t) \ dt = \frac{1}{a^2} \int^x \frac{dv}{v^2}$$

   $$\text{with } \Biggl \{ \begin{gather*} v =\sin(u) \\ dv = \cos(u) du \end{gather*}$$

   We are in the case of a standard antiderivative :

   $$ \forall x \in \mathbb{R}, \ \int^x \frac{dt}{t^2} = - \frac{1}{x} $$

   So in our case:

   $$ \int^x J_3(t) \ dt = - \frac{1}{a^2} \frac{1}{v} $$

   Now going back the variable changes in their order of assignment.

   $$ \int^x J_3(t) \ dt = - \frac{1}{a^2} \frac{1}{\sin(u)} $$
   $$ \int^x J_3(t) \ dt = - \frac{1}{a^2} \frac{1}{\sin\left(\operatorname{Arctan}\left(\frac{x}{|a|}\right)\right)} $$

   But, we saw above that:

   $$\sec^2(\operatorname{Arctan}(x)) = 1+ x^2 \hspace{4em} \Big[\sec^2(\operatorname{Arctan}) \Bigr]$$

   So,

   $$\frac{1}{\cos^2(\operatorname{Arctan}(x))} = 1+ x^2 $$

   $$\frac{1}{1 - \sin^2(\operatorname{Arctan}(x))} = 1+ x^2 $$

   $$\frac{1}{1+ x^2} = 1 - \sin^2(\operatorname{Arctan}(x)) $$

   $$\sin^2(\operatorname{Arctan}(x)) = 1 - \frac{1}{1+ x^2} $$

   $$\sin^2(\operatorname{Arctan}(x)) = \sqrt{ \frac{ 1 + x^2 -1}{1+ x^2} } $$

   $$\sin(\operatorname{Arctan}(x)) = \frac{x}{\sqrt{1+ x^2}} \hspace{4em} \Big[\sin(\operatorname{Arctan}) \Bigr]$$

   So, replacing in the previous expression:

   $$ \int^x J_3(t) \ dt = - \frac{1}{a^2} \frac{\sqrt{1+ \left(\frac{x}{|a|}\right)^2}}{\left(\frac{x}{|a|}\right)} $$
   $$ \int^x J_3(t) \ dt = - \frac{|a|}{|a|^2} \frac{\sqrt{1+ \left(\frac{x}{a}\right)^2}}{x} $$
   $$ \int^x J_3(t) \ dt = - \frac{1}{a^2} \frac{\sqrt{a^2+x^2}}{x} $$

   And as a result,

   $$ \forall a \in \hspace{0.04em} \mathbb{R}^+, \ \forall x \in \hspace{0.04em} \mathbb{R}^*, $$
   $$ \int^x \frac{dt}{t^2\sqrt{a^2 + t^2}} =- \frac{1}{a^2} \frac{\sqrt{a^2+x^2}}{x}$$

4. Simple root\(: \sqrt{a^2 + t^2} \)

   $$ \forall (a, x) \in \hspace{0.04em} \mathbb{R}^2, \ J_4(x) = \sqrt{a^2 + t^2} $$
   $$ \int^x J_4(t) \ dt = \int^x \sqrt{a^2 + t^2} \ dt $$

   We can set down \(t = |a| \ \tan(u)\).

   $$ \int^x J_4(t) \ dt = \int^x \sqrt{a^2 + |a|^2 \tan^2(u)} \ |a| \ \sec^2(u) \ du $$

   $$\text{with } \Biggl \{ \begin{gather*} t = |a| \ \tan(u) \\ dt = |a| \ (1 + \tan^2(u)) \ du = |a| \ \sec^2(u) \ du \end{gather*}$$
   $$ \int^x J_4(t) \ dt = a^2 \int^x \sqrt{\sec^2(u)} \ \sec^2(u) \ du $$
   $$ \int^x J_4(t) \ dt = a^2 \int^x \frac{\sec(u)}{|\sec(u)|} \sec^3(u) \ du $$

   We saw above that this sign generator always worth \(1\):

   $$ \forall x \in \hspace{0.04em} \mathbb{R}^*, \ \frac{\sec(u)}{|\sec(u)|} = 1 $$

   Consequently, removing it from the integrand:

   $$ \int^x J_4(t) \ dt = a^2 \int^x \sec^3(u) \ du \qquad(J_4) $$
   $$ \int^x J_4(t) \ dt = a^2 \int^x \sec(u)\tan'(u) \ du $$

   On peut alors faire an integration by parts :

   $$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[\sec(u)\tan(u)\Bigr]^u - \int^x \sec'(u)\tan(u) \ du \Biggr)$$
   $$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[\sec(u)\tan(u)\Bigr]^u - \int^x \sec(u)\tan^2(u) \ du \Biggr) \hspace{3em} \Bigl(\text{with } \sec'(u) = \sec(u)\tan(u)\Bigr)$$
   $$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[\sec(u)\tan(u)\Bigr]^u - \int^x \sec(u)(\sec^2(u)-1) \ du \Biggr) \hspace{3em} \Bigl(\text{with } \tan^2(u) = \sec^2(u)-1\Bigr) $$
   $$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[\sec(u)\tan(u)\Bigr]^u - \int^x \Bigl( \sec^3(u) - \sec(u) \Bigr) \ du \Biggr)$$
   $$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[\sec(u)\tan(u)\Bigr]^u - \int^x \sec^3(u) \ du + \int^x \sec(u) \ du \Biggr) $$

   Now, with the previous expression \((J_4)\), we had:

   $$ \int^x J_4(t) \ dt = a^2 \int^x \sec^3(u) \ du \qquad(J_4) $$

   Then, replacing it:

   $$ \int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[\sec(u)\tan(u)\Bigr]^u + \int^x \sec(u) \ du \Biggr) - \int^x J_4(t) \ dt $$
   $$ 2\int^x J_4(t) \ dt = a^2 \Biggl( \Bigl[\sec(u)\tan(u)\Bigr]^u + \int^x \sec(u) \ du \Biggr) $$
   $$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl( \Bigl[\sec(u)\tan(u)\Bigr]^u + \int^x \sec(u) \ du \Biggr) $$

   And as we know the antiderivative of \(\sec(x)\) , we can replace it.

   $$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl( \Bigl[\sec(u)\tan(u)\Bigr]^u + \Bigl[\ln \left|\sec(u) + \tan(u) \right|\Bigr]^u \Biggr) \qquad(J_4^*) $$

   Let us finally replace \(u\) by its value:

   $$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl( \sec\left(\operatorname{Arctan}\left( \frac{x}{|a|} \right) \right) \tan\left(\operatorname{Arctan}\left( \frac{x}{|a|} \right)\right) + \ln\left| \sec\left(\operatorname{Arctan}\left( \frac{x}{|a|} \right) \right) + \tan\left(\operatorname{Arctan}\left( \frac{x}{|a|} \right)\right) \right| \Biggr) $$

   But, we saw above that:

   $$\sec^2(\operatorname{Arctan}(x)) = 1+ x^2 \Longleftrightarrow \sec\left(\operatorname{Arctan}(x)\right) = \sqrt{ 1 + x^2 } \hspace{4em} \Big[\sec^2(\operatorname{Arctan}) \Bigr]$$

   So,

   $$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl( \sqrt{ 1 + \left(\frac{x}{|a|} \right)^2 } \ \frac{x}{|a|} + \ln\left| \sqrt{ 1 + \left(\frac{x}{|a|} \right)^2 } + \frac{x}{|a|} \right| \Biggr) $$
   $$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl(\frac{x}{a^2}\sqrt{a^2 +x^2} + \ln\left| \frac{1}{|a|} \sqrt{a^2 +x^2} + \frac{x}{|a|} \right| \Biggr) $$
   $$ \int^x J_4(t) \ dt = \frac{a^2}{2} \Biggl(\frac{x}{a^2}\sqrt{a^2 +x^2} + \ln\left| \frac{1}{|a|} \Bigl( \sqrt{a^2 +x^2} + x \Bigr) \right| \Biggr) $$
   $$ \int^x J_4(t) \ dt = \frac{1}{2} \Biggl(x \ \sqrt{a^2 +x^2} + a^2 \ \ln\left|\sqrt{a^2 +x^2} + x \right| - \ln\Bigl| |a| \Bigr| \Biggr) $$

   The constant \(- \ln\Bigl| |a| \Bigr|\) will be absorbed by the main constant integration and:

   $$ \Big[ \forall a \in \hspace{0.04em} \mathbb{R}^*, \ \forall x \in \mathbb{R}\Bigr] \lor \Big[ (a = 0) \land (x > 0) \Bigr], $$
   $$ \int^x \sqrt{a^2 + t^2} \ dt = \frac{x \ \sqrt{a^2 +x^2} + a^2 \ \ln\left|\sqrt{a^2 +x^2} + x \right|}{2} $$

Integrals containing \(\sqrt{t^2 - a^2}\)

1. Fraction with a root at the denominator\(: \frac{1}{\sqrt{t^2 - a^2}}\)

   $$ \forall a \in \hspace{0.04em} \mathbb{R}^*, \ \forall x \in \Bigl] -\infty, -|a| \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]|a|, +\infty \Bigr[, \ K_1(x) = \frac{1}{\sqrt{x^2 - a^2}} $$
   $$ \int^x K_1(t) \ dt = \int^x \frac{dt}{\sqrt{t^2 - a^2}}$$

   1. Setting down \( t = |a| \ \cosh(u) \)

      By setting down \( t = |a| \ \cosh(u) \), we do have:

      $$ \int^x K_1(t) \ dt = \int^x \frac{|a| \ \sinh(u) \ du}{\sqrt{|a|^2 \ \cosh^2(u) - a^2}} $$

      $$\text{with } \Biggl \{ \begin{gather*} t = |a| \ \cosh(u) \\ dt = |a| \ \sinh(u) \ du \end{gather*}$$
      $$ \int^x K_1(t) \ dt = \frac{|a|}{|a|}\int^x \frac{\sinh(u) \ du}{\sqrt{ \sinh^2(u)}} $$
      $$ \int^x K_1(t) \ dt = \int^x \frac{\sinh(u) \ du}{|\sinh(u)|} \qquad \left(\text{with } u \neq 0 \Longleftrightarrow t \neq |a|, \text{ which is the case} \right) $$

      The \(\operatorname{Argcosh}\) function being always positive, so do the \(\sinh(\operatorname{Argcosh})\) one, and :

      $$ \forall a \in \hspace{0.04em} \mathbb{R}^+, \ \forall x \in \Bigl] -\infty, -|a| \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]|a|, +\infty \Bigr[, \ \frac{\sinh(u)}{|\sinh(u)|} = 1$$

      So, we now integrate only :

      $$ \int^x K_1(t) \ dt = \int^x du $$
      $$ \int^x K_1(t) \ dt = \Bigr[ u \Bigr]^{u=\operatorname{Argcosh} \left( \frac{x}{|a|} \right)}$$

      And finally,

      $$ \forall a \in \hspace{0.04em} \mathbb{R}^+, \ \forall x \in \hspace{0.04em} ]a, +\infty[, $$

      $$ \int^x \frac{dt}{\sqrt{t^2 - a^2}} = \operatorname{Argcosh}\left( \frac{x}{|a|}\right) $$

      $$(6)$$

   2. Setting down \( t = |a| \ \sec(u) \)

      Moreover, by setting down \( t = |a| \ \sec(u) \), we do have:

      $$ \int^x K_1(t) \ dt = \int^x \frac{|a| \ \sec(u)\tan(u) \ du }{\sqrt{|a|^2 \ \sec^2(u) - a^2}}$$

      $$\text{with } \Biggl \{ \begin{gather*} t = |a| \ \sec(u) \\ dt = |a| \ \sec^2(u)\sin(u) \ du = |a| \ \sec(u)\tan(u) \ du \end{gather*}$$
      $$ \int^x K_1(t) \ dt = \frac{|a|}{|a|}\int^x \frac{\sec(u)\tan(u) \ du }{\sqrt{ \sec^2(u) -1}} $$
      $$ \int^x K_1(t) \ dt = \int^x \frac{\sec(u)\tan(u) \ du }{\sqrt{ \tan^2(u)}} $$
      $$ \int^x K_1(t) \ dt = \int^x \frac{\sec(u)\tan(u) \ du }{|\tan(u)|} $$

      Let us calculate the value of this sign generator placed under the integrand.

      Using again the derivatives of reciprocal functions , we do have:

      $$ \left( \operatorname{Arcsec}(x) \right)' = \frac{1}{ \sec\left(\operatorname{Arcsec}(x)\right) \ \tan\left(\operatorname{Arcsec}(x)\right)} $$

      $$\text{with } \Biggl \{ \begin{gather*} f^{-1} = \operatorname{Arcsec}(x) \\ f = \sec(x) \Longrightarrow f' = \sec(x)\tan(x) \end{gather*}$$

      $$ \Longleftrightarrow \ \tan(\operatorname{Arcsec}(x)) = \frac{1}{ x} \times \frac{1}{ (\operatorname{Arcsec}(x) (x))' } $$

      But, we know the derivative of the \(\operatorname{Arcsec}(x)\) function :

      $$ \forall x \in \hspace{0.04em} ]-\infty, \hspace{0.1em} -1[\hspace{0.1em}\cup \hspace{0.1em}]1, \hspace{0.1em} +\infty[, $$

      $$ \operatorname{Arcsec}(x)' = \frac{1}{ x^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ x^2}}} $$

      $$ \operatorname{Arcsec}(x)' = \frac{1}{|x|} \times \frac{1}{ \sqrt{ x^2 - 1}} $$

      By combining the two previous expressions, we obtain:

      $$\tan(\operatorname{Arcsec}(x)) = \frac{x}{|x|} \times \sqrt{ x^2 - 1 } \qquad \Big[\tan(\operatorname{Arcsec}(x))\Bigr]$$
      $$\frac{\tan(u)}{|\tan(u)|} = \frac{ \frac{\ \frac{x}{|a|} }{\left| \frac{x}{|a|} \right|} \times \sqrt{ \left( \frac{x}{|a|} \right)^2 - 1 } }{ \left| \frac{\ \frac{x}{|a|} }{\left| \frac{x}{|a|} \right|} \times \sqrt{ \left( \frac{x}{|a|} \right)^2 - 1 } \right| }$$
      $$\frac{\tan(u)}{|\tan(u)|} = \frac{ \frac{ x }{|x|} \times \sqrt{ \frac{x^2}{a^2} - 1} }{ \left| \frac{ x }{|x|} \times \sqrt{ \frac{x^2}{a^2} - 1} \right| }$$
      $$\frac{\tan(u)}{|\tan(u)|} = \frac{ x }{|x|} $$

      In the study interval, this ratio is worth \(\pm 1\):

      $$ \frac{\tan(u)}{|\tan(u)|} = \Biggl \{ \begin{gather*} = -1, \ \forall x \in \hspace{0.04em} \bigl] -\infty, -a \bigr[ \\ = 1, \ \forall x \in \hspace{0.04em} ]a, +\infty [ \end{gather*} $$

      We now integrate:

      $$ \int^x K_1(t) \ dt = \frac{\tan(u)}{|\tan(u)|} \int^x \sec(u) \ du $$

      Let us calculate the value of \( {\displaystyle \int^x} \sec(u) \ du\) before to take care of the sign:

      $$ \int^x \sec(u) \ du = \Bigl[ \ln \left| \sec(x) + \tan(x) \right| \Bigr]^{u = \operatorname{Arcsec}(x)\left( \frac{x}{|a|} \right)} $$
      $$ \int^x \sec(u) \ du = \ln \left| \frac{x}{|a|} + \tan\left(\operatorname{Arcsec}(x)\left( \frac{x}{|a|} \right) \right) \right| $$

      Using again \(\Big[ \tan(\operatorname{Arcsec}(x))\Bigr] \).

      $$\tan(\operatorname{Arcsec}(x)) = \frac{x}{|x|} \times \sqrt{ x^2 - 1 } \qquad \Big[\tan(\operatorname{Arcsec}(x))\Bigr]$$

      We can now inject the last expression into our previous expression:

      $$ \int^x \sec(u) \ du = \ln \left| \frac{x}{|a|} + \frac{\frac{x}{|a|}}{\left|\frac{x}{|a|}\right|} \sqrt{ \left( \frac{x}{|a|} \right)^2 - 1 } \right| $$
      $$ \int^x \sec(u) \ du = \ln \left| \frac{x}{|a|} + \frac{x}{|x|}\frac{ \sqrt{ x^2 - a^2 }}{ |a|} \right| $$
      $$ \int^x \sec(u) \ du = \ln \left| \frac{1}{|a|} \left( x + \frac{x\sqrt{ x^2 - a^2 }}{|x|} \right) \right| $$
      $$ \int^x \sec(u) \ du = \ln \left| \frac{x\sqrt{ x^2 - a^2 }}{|x|} + x \right| -\ln|a| $$

      The constant \(-\ln|a| \) being absorbed by the main integration constant, we finally obtain that:

      So the final integral is worth:

      $$ \int^x K_1(t) \ dt = \frac{\tan(u)}{|\tan(u)|} \ \ln \left| \frac{x\sqrt{ x^2 - a^2 }}{|x|} + x \right| $$
      $$ \left(\text{with } \frac{\tan(u)}{|\tan(u)|} = \Biggl \{ \begin{gather*} = -1, \ \forall x \in \hspace{0.04em} \bigl] -\infty, -a \bigr[ \\ = 1, \ \forall x \in \hspace{0.04em} ]a, +\infty [ \end{gather*} \right)$$

      We then have to manage the two cases:

      1. For the negative part: \(\bigl] -\infty, -a \bigr[ \)

         \(x\) is negative and the sign generator too:

         $$ \forall x \in \hspace{0.04em} \bigl] -\infty, -a \bigr[, \ \int^x K_1(t) \ dt = -\ln \left|- \sqrt{ x^2 - a^2 } - |x| \right| $$
         $$ \int^x K_1(t) \ dt = -\ln \left| \sqrt{ x^2 - a^2 } + |x| \right| $$
         $$ \int^x K_1(t) \ dt = \ln \left( \frac{1}{\left| \sqrt{ x^2 - a^2 } + |x| \right|} \right) $$

         We multiply both numerator and denominator by the conjugate of the denominator :

         $$ \int^x K_1(t) \ dt = \ln \left( \frac{1 \textcolor{rgb(118 139 240)}{\times \left| \sqrt{ x^2 - a^2 } - |x| \right|}}{\left| \sqrt{ x^2 - a^2 } + |x| \right| \textcolor{rgb(118 139 240)}{\times \left| \sqrt{ x^2 - a^2 } - |x| \right|} } \right) $$
         $$ \int^x K_1(t) \ dt = \ln \left( \frac{\left| \sqrt{ x^2 - a^2 } - |x| \right|}{\left|x^2 - a^2 - x^2 \right|} \right) $$
         $$ \int^x K_1(t) \ dt = \ln \left( \frac{\left| \sqrt{ x^2 - a^2 } - |x| \right|}{\left| -a^2 \right|} \right) $$
         $$ \int^x K_1(t) \ dt = \ln \left( \frac{ \left| \sqrt{ x^2 - a^2 } - |x| \right| }{a^2} \right) $$
         $$ \int^x K_1(t) \ dt = \ln \left| \sqrt{ x^2 - a^2 } - |x| \right| - \ln(a^2) $$
         $$ \int^x K_1(t) \ dt = \ln \left| \sqrt{ x^2 - a^2 } + x \right| - 2 \ \ln(a) $$

         The constant \(- 2 \ \ln(a)\) will be also absorbed by the main integration constant.

      2. For the positive part: \(] a, +\infty [ \)

         \(x\) is positive and the sign generator too:

         $$ \forall x \in \hspace{0.04em} ] a, +\infty [, \ \int^x K_1(t) \ dt = \ln \left|\sqrt{ x^2 - a^2 } + x \right| $$

      3. Conclusion

         In any case \((x < -a) \ or \ (x > a)\), the integral is worth:

         $$ \Bigl[ \forall a \in \hspace{0.04em} \mathbb{R}^*, \ \forall x \in \Bigl] -\infty, -|a| \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]|a|, +\infty \Bigr[ \lor \Bigl[ (x = 0) \land (x > 0) \Bigr], $$

         $$\int^x \frac{dt}{\sqrt{t^2 - a^2}} = \ln \left| \sqrt{ x^2 - a^2 } + x \right| $$

         $$(7)$$

   Both expressions \((6)\) and \((7)\) having a common member, they are equal up to a constant in their common interval.

   $$ \operatorname{Argcosh}\left( \frac{x}{a}\right) + C = \ln \left| \sqrt{ x^2 - a^2 } + x \right| $$

   Let us determine this constant by taking a value of \(x\), for example \(x = 0\).

   $$ \operatorname{Argcosh}\left( \frac{1}{|a|}\right) = 0 $$

   $$\ln \left| \sqrt{1 - a^2 } + 1 \right|$$

   Then, we find that:

   $$ \left(C_1 = \ln \left| \sqrt{1^2 - a^2 } + 1 \right| + C_2 \right) \Longleftrightarrow \left(C_1 - C_2 = \ln \left| \sqrt{1 - a^2 } + 1 \right| \right)$$
   $$ \operatorname{Argcosh}\left( \frac{x}{|a|}\right) + \ln \left| \sqrt{1 - a^2 } + 1 \right| = \ln \left| \sqrt{ x^2 - a^2 } + x \right| $$

   We then have as a bonus an explicit definition of the \(\operatorname{Argcosh}\) function with \(a = 1\):

   $$ \forall x \in [1, +\infty[,$$
   $$ \operatorname{Argcosh}(x) = \ln \left| x + \sqrt{ x^2 - 1} \right| $$

2. Fraction with \(t\) and a root in the denominator\(: \frac{1}{t\sqrt{t^2 - a^2}}\)

   $$\forall a \in \hspace{0.04em} \mathbb{R}^*, \ \forall x \in \Bigl] -\infty, -|a| \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]|a|, +\infty \Bigr[, \ K_2(x) = \frac{1}{x\sqrt{x^2 - a^2 }} $$
   $$ \int^x K_2(t) \ dt = \int^x \frac{dt}{t\sqrt{t^2 - a^2}} $$

   As before, we set down: \(u = \sqrt{t^2 - a^2}\).

   $$ \int^x K_2(t) \ dt = \int^x \frac{du}{t^2} $$

   $$\text{with } \Biggl \{ \begin{gather*} u = \sqrt{t^2 - a^2} \\ du = \frac{t}{\sqrt{t^2 - a^2}} \ dt \end{gather*}$$

   But,

   $$u = \sqrt{a^2 + t^2} \Longleftrightarrow t^2 = u^2 + a^2$$

   We now integrate:

   $$ \int^x K_2(t) \ dt = \int^x \frac{du}{u^2 + a^2} $$

   We then recognize a standard antiderivative .

   $$ \int^x K_2(t) \ dt = \frac{1}{a^2} \int^x \frac{du}{ \left(\frac{u}{a} \right)^2 + 1 } $$
   $$ \int^x K_2(t) \ dt = \frac{1}{a} \Biggl[ \operatorname{Arctan}\left(\frac{u}{a}\right) \Biggr]^{u= \sqrt{x^2-a^2}} $$
   $$ \int^x K_2(t) \ dt = \frac{1}{a} \operatorname{Arctan} \left(\frac{\sqrt{x^2 - a^2}}{a} \right) $$

   And as a result,

   $$ \forall a \in \hspace{0.04em} \mathbb{R}^+, \ \forall x \in \Bigl] -\infty, -|a| \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]|a|, +\infty \Bigr[,$$
   $$ \int^x \frac{dt}{t\sqrt{t^2 - a^2}}dt = \frac{1}{a} \operatorname{Arctan} \left(\frac{\sqrt{x^2 - a^2}}{a} \right) $$

3. Fraction with \(t^2\) and a root in the denominator\(: \frac{1}{t^2\sqrt{t^2 - a^2}}\)

   $$\forall a \in \hspace{0.04em} \mathbb{R}^*, \ \forall x \in \Bigl] -\infty, -|a| \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]|a|, +\infty \Bigr[, \ K_3(x) = \frac{1}{x^2\sqrt{x^2 - a^2}} $$
   $$ \int^x K_3(t) \ dt = \int^x \frac{dt}{t^2\sqrt{t^2 - a^2}} $$

   We set down a new variable: \(t = |a| \ \cosh(u)\).

   $$ \int^x K_3(t) \ dt = \int^x \frac{ |a| \ \sinh(u) \ du }{ |a|^2 \ \cosh^2(u)\sqrt{|a|^2 \ \cosh^2(u) - a^2}} $$

   $$\text{with } \Biggl \{ \begin{gather*} t = |a| \ \cosh(u) \\ dt = |a| \ \sinh(u) \ du \end{gather*}$$

   And consequently,

   $$ \int^x K_3(t) \ dt = \frac{1}{a^2} \int^x \frac{ \sinh(u) }{ \cosh^2(u)\sqrt{\sinh^2(u)}} \ du $$
   $$ \int^x K_3(t) \ dt = \frac{1}{a^2} \int^x \operatorname{sech}^2(u) \frac{ \sinh(u) }{ |\sinh(u)|} \ du $$

   We already calculated it above and this sign generator always worth \(1\):

   $$ \forall a \in \hspace{0.04em} \mathbb{R}^+, \ \forall x \in \Bigl] -\infty, -|a| \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]|a|, +\infty \Bigr[, \ \frac{\sinh(u)}{|\sinh(u)|} = 1$$

   So we now integrate:

   $$ \int^x K_3(t) \ dt = \frac{1}{a^2} \int^x \operatorname{sech}^2(u) \ du $$

   We then in a case of a standard antiderivative :

   $$ \forall x \in \mathbb{R}, \ \int^x \operatorname{sech}^2(t) \ dt = \tanh(t) $$

   Now, replacing it we obtain:

   $$ \int^x K_3(t) \ dt = \frac{1}{a^2} \tanh\left(\operatorname{Argcosh}\left( \frac{x}{|a|} \right) \right) $$
   $$ \int^x K_3(t) \ dt = \frac{1}{a^2} \frac{\sinh\left(\operatorname{Argcosh}\left( \frac{x}{|a|} \right) \right)}{\cosh\left(\operatorname{Argcosh}\left( \frac{x}{|a|} \right) \right)}$$

   Using again the derivatives of reciprocal functions , we do have:

   $$ \left( \operatorname{Argcosh}(x) \right)' = \frac{1}{ \sinh\left(\operatorname{Argcosh}(x)\right)} $$

   $$\text{with } \Biggl \{ \begin{gather*} f^{-1} = \operatorname{Argcosh}(x) \\ f = \cosh(x) \Longrightarrow f' = \sinh(x) \end{gather*}$$

   $$ \Longleftrightarrow \ \sinh\left(\operatorname{Argcosh}(x)\right) = \frac{1}{ \left( \operatorname{Argcosh}(x) \right)' } $$

   But we know the derivative of the \(\operatorname{Argcosh}(x)\) function :

   $$ \forall x \in \hspace{0.04em} ]1, \hspace{0.1em} +\infty[, $$

   $$ \operatorname{Argcosh}(x)' = \frac{1}{\sqrt{x^2 -1}} $$

   En combinant les deux expressions précédentes, on obtient :

   $$\sinh\left(\operatorname{Argcosh}(x)\right) = x^2 - 1 \hspace{4em} \Big[\sinh(\operatorname{Argcosh}) \Bigr]$$

   So, replacing it:

   $$ \int^x K_3(t) \ dt = \frac{1}{a^2} \frac{ \sqrt{ \left( \frac{x}{|a|} \right)^2 - 1 }}{ \left( \frac{x}{|a|} \right) }$$
   $$ \int^x K_3(t) \ dt = \frac{|a|}{a^2 |a|} \frac{\sqrt{ x^2 - a^2 } }{x}$$

   As a result we obtain,

   $$\ \forall a \in \hspace{0.04em} \mathbb{R}^+, \ \forall x \in \Bigl] -\infty, -|a| \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]|a|, +\infty \Bigr[, $$
   $$ \int^x \frac{dt}{t^2\sqrt{t^2 - a^2}} =\frac{1}{a^2} \frac{\sqrt{ x^2 - a^2 } }{x}$$

4. Simple root\(: \sqrt{t^2 - a^2} \)

   $$\forall a \in \mathbb{R}, \ \forall x \in \Bigl] -\infty, -|a| \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]|a|, +\infty \Bigr[, \ K_4(x) = \sqrt{t^2 - a^2} $$
   $$ \int^x K_4(t) \ dt = \int^x \sqrt{t^2 - a^2}\ dt $$

   We can set down the variable: \(t = |a| \ \sec(u)\) :

   $$ \int^x K_4(t) \ dt = \int^x \sqrt{|a|^2 \sec^2(u) - a^2} \ |a| \ \sec(u)\tan(u) \ du $$

   $$\text{with } \Biggl \{ \begin{gather*} t = |a| \ \sec(u) \\ dt = |a| \ \sec(u)\tan(u) \ du \end{gather*}$$
   $$ \int^x K_4(t) \ dt = a^2 \int^x \sqrt{\sec^2(u) - 1} \ \sec(u)\tan(u) \ du$$
   $$ \int^x K_4(t) \ dt = a^2 \int^x \sqrt{\tan^2(u)} \ \sec(u)\tan(u) \ du$$
   $$ \int^x K_4(t) \ dt = a^2 \int^x \frac{\tan(u)}{|\tan(u)|} \ \sec(u) \ \tan^2(u) \ du$$

   We saw above that this sign generator was worth \(\pm 1\):

   $$ \frac{\tan(u)}{|\tan(u)|} = \Biggl \{ \begin{gather*} = -1, \ \forall x \in \hspace{0.04em} \bigl] -\infty, -a \bigr[ \\ = 1, \ \forall x \in \hspace{0.04em} ]a, +\infty [ \end{gather*} $$

   And we now integrate:

   $$ \int^x K_4(t) \ dt = a^2 \frac{\tan(u)}{|\tan(u)|} \int^x \sec(u) \ \tan^2(u) \ du$$

   Let us firstly calculate the value of the integral \(a^2 {\displaystyle \int^x} \sec(u) \ \tan^2(u) \ du \) before to take care of the sign:

   $$ a^2 \int^x \sec(u) \ \tan^2(u) \ du = a^2 \int^x \sec(u)(\sec^2(u) -1)u \ du$$
   $$ a^2 \int^x \sec(u) \ \tan^2(u) \ du = a^2 \int^x \sec^3(u) \ du - a^2 \int^x \sec(u) \ du$$

   These two integrals have already been calculated:

   The first on this page:

   $$\int^x \sec(u) \ du = \ln \left|\sec(x) + \tan(x) \right|$$

   And the second one previously:

   $$ \int^x \sec^3(u) \ du = \frac{1}{2} \Biggl( \Bigl[\sec(u)\tan(u)\Bigr]^u + \Bigl[\ln \left|\sec(u) + \tan(u) \right|\Bigr]^u \Biggr) \qquad(J_3^*) $$

   So,

   $$ a^2 \int^x \sec(u) \ \tan^2(u) \ du = \frac{a^2}{2} \biggl[ \sec(u)\tan(u) + \ln \left|\sec(u) + \tan(u) \right| \biggr]^u - a^2 \biggl[ \ln \left|\sec(u) + \tan(u) \right| \biggr]^u \ du $$
   $$ a^2 \int^x \sec(u) \ \tan^2(u) \ du = \frac{a^2}{2} \Biggl[ \sec\left( \operatorname{Arcsec}(x)\left(\frac{x}{|a|} \right) \right)\tan\left( \operatorname{Arcsec}(x)\left(\frac{x}{|a|} \right) \right) + \ln \Biggl| \ \sec\left( \operatorname{Arcsec}(x)\left(\frac{x}{|a|} \right) \right) + \tan\left( \operatorname{Arcsec}(x)\left(\frac{x}{|a|} \right) \right) \ \Biggr| \Biggr] - a^2 \ \ln \Biggl| \ \sec\left( \operatorname{Arcsec}(x)\left(\frac{x}{|a|} \right) \right) + \tan\left( \operatorname{Arcsec}(x)\left(\frac{x}{|a|} \right) \right) \ \Biggr| $$
   $$ a^2 \int^x \sec(u) \ \tan^2(u) \ du = \frac{a^2}{2} \Biggl[ \frac{x}{|a|} \tan\left( \operatorname{Arcsec}(x)\left(\frac{x}{|a|} \right) \right) + \ln \Biggl| \ \frac{x}{|a|} + \tan\left( \operatorname{Arcsec}(x)\left(\frac{x}{|a|} \right) \right) \ \Biggr| \Biggr] - a^2 \ \ln \Biggl| \ \frac{x}{|a|} + \tan\left( \operatorname{Arcsec}(x)\left(\frac{x}{|a|} \right) \right) \ \Biggr| $$

   Moreover, we have also seen that:

   $$\tan(\operatorname{Arcsec}(x)) = \frac{x}{|x|} \times \sqrt{ x^2 - 1 } \qquad \Big[\tan(\operatorname{Arcsec}(x))\Bigr]$$

   The replacing it we do have now:

   $$ a^2 \int^x \sec(u) \ \tan^2(u) \ du = \frac{a^2}{2} \Biggl[ \frac{x}{|a|} \frac{\frac{x}{|a|}}{\left|\frac{x}{|a|}\right|} \sqrt{\left(\frac{x}{|a|} \right)^2 - 1} + \ln \Biggl| \ \frac{x}{|a|} + \frac{\frac{x}{|a|}}{\left|\frac{x}{|a|}\right|} \sqrt{\left(\frac{x}{|a|} \right)^2 - 1} \ \Biggr| \Biggr] - a^2 \ \ln \Biggl| \ \frac{x}{|a|} + \frac{\frac{x}{|a|}}{\left|\frac{x}{|a|}\right|} \sqrt{\left(\frac{x}{|a|} \right)^2 - 1} \ \Biggr| $$
   $$ a^2 \int^x \sec(u) \ \tan^2(u) \ du = \frac{a^2}{2} \Biggl[ \frac{x^2}{a^2|x|} \sqrt{x^2 - a^2} + \ln \Biggl| \ \frac{x}{|a|} + \frac{x}{|ax|} \sqrt{x^2 - a^2} \ \Biggr| \Biggr] - a^2 \ \ln \Biggl| \ \frac{x}{|a|} + \frac{x}{|ax|} \sqrt{x^2 - a^2} \ \Biggr| $$
   $$ a^2 \int^x \sec(u) \ \tan^2(u) \ du = \frac{a^2}{2} \Biggl[ \frac{x^2}{a^2|x|} \sqrt{x^2 - a^2} + \ln \Biggl| \frac{1}{|a|} \left( x + \frac{x \sqrt{x^2 - a^2}}{|x|} \right) \ \Biggr| \Biggr] - a^2 \ \ln \Biggl| \frac{1}{|a|} \left( x + \frac{x \sqrt{x^2 - a^2}}{|x|} \right) \ \Biggr| $$

   We can now factorize by the big factor containing \(\ln|X|\):

   $$ a^2 \int^x \sec(u) \ \tan^2(u) \ du = \frac{1}{2} \frac{x^2}{|x|}\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl| \frac{1}{|a|} \left( x + \frac{x \sqrt{x^2 - a^2}}{|x|} \right) \ \Biggr| $$
   $$ a^2 \int^x \sec(u) \ \tan^2(u) \ du = \frac{1}{2}\frac{x^2}{|x|}\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl| x + \frac{x \sqrt{x^2 - a^2}}{|x|} \ \Biggr| + \frac{ \ln|a| a^2}{2} $$

   The constant \( + \frac{ \ln|a| a^2}{2} \) being absorbed by the main integration constant, we finally obtain by taking in consideration the sign generator left aside:

   $$ \int^x K_4(t) \ dt = \frac{\tan(u)}{|\tan(u)|} \Biggl( \frac{1}{2}\frac{x^2}{|x|}\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl| x + \frac{x \sqrt{x^2 - a^2}}{|x|} \ \Biggr| \Biggr) $$
   $$ \left(\text{with } \frac{\tan(u)}{|\tan(u)|} = \Biggl \{ \begin{gather*} = -1, \ \forall x \in \hspace{0.04em} \bigl] -\infty, -a \bigr[ \\ = 1, \ \forall x \in \hspace{0.04em} ]a, +\infty [ \end{gather*} \right)$$

   We then have to cases to manage now:

   1. For the negative part: \(\bigl] -\infty, -a \bigr[ \)

      \(x\) is negative and the sign generator too, so:

      $$ \forall x \in \hspace{0.04em} \bigl] -\infty, -a \bigr[, \ \int^x K_4(t) \ dt = -\Biggl( -\frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl| -|x| - \sqrt{x^2 - a^2} \ \Biggr| \Biggr) $$
      $$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} + \frac{a^2}{2} \ln \Biggl| -|x| - \sqrt{x^2 - a^2} \ \Biggr| $$
      $$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl( \frac{1}{\left| -|x| - \sqrt{x^2 - a^2} \right|} \ \Biggr) $$

      We multiply both numerator and denominator by the conjugate of the denominator :

      $$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl( \frac{1 \textcolor{rgb(118 139 240)}{ \times \left| -|x| + \sqrt{x^2 - a^2} \right|} }{\left| -|x| - \sqrt{x^2 - a^2} \right| \textcolor{rgb(118 139 240)}{ \times \left| -|x| + \sqrt{x^2 - a^2} \right|} } \ \Biggr) $$
      $$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl( \frac{\left| -|x| + \sqrt{x^2 - a^2} \right|}{\left| \left(-|x|\right)^2 - \left(x^2 - a^2\right) \right|} \ \Biggr) $$
      $$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl( \frac{\left| -|x| + \sqrt{x^2 - a^2} \right|}{\left| x^2 - x^2 + a^2 \right|} \ \Biggr) $$
      $$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl( \frac{\left| -|x| + \sqrt{x^2 - a^2} \right|}{\left| a^2 \right|} \ \Biggr) $$
      $$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl( \left| \frac{-|x| + \sqrt{x^2 - a^2}}{a^2} \right| \ \Biggr) $$
      $$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl| -|x| + \sqrt{x^2 - a^2} \Biggr| +\ln(a^2) $$
      $$ \int^x K_4(t) \ dt = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl| x + \sqrt{x^2 - a^2} \Biggr| +2 \ \ln(a) $$

      The constant \( +2 \ \ln(a) \) will be also absorbed by the main integration constant.

   2. For the positive part: \(] a, +\infty [ \)

      \(x\) is positive and the sign generator too, so:

      $$ \forall x \in \hspace{0.04em} ] a, +\infty [, \ \int^x K_4(t) \ dt =\frac{1}{2}x\sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \Biggl| x+ \sqrt{x^2 - a^2} \ \Biggr| $$

   3. Conclusion

      In any case, \((x < -a) \ or \ (x > a)\), this integral is worth:

      $$ \Bigl[ \forall a \in \hspace{0.04em} \mathbb{R}^*, \ \forall x \in \Bigl] -\infty, -|a| \Bigr[ \hspace{0.04em} \cup \hspace{0.04em} \Bigl]|a|, +\infty \Bigr[ \lor \Bigl[ (x = 0) \land (x > 0) \Bigr],$$
      $$ \int^x \sqrt{t^2 - a^2} \ dt = \frac{ x\sqrt{x^2 - a^2} - a^2 \ \ln \left| x+ \sqrt{x^2 - a^2} \ \right| }{2} $$

Integration methods and antiderivatives recap table