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Standard antiderivatives and general integration methods

Since each antiderivative of a given function is equal up to a constant, we will use the unified notation \({\displaystyle \int^x} f(t) \ dt \),meaning this family of primitives (or general antiderivative ).

We will sometimes use the capital letter of a function to signify its antiderivative .

Recall on standard antiderivatives

A certain amount of antiderivative can be directly calculated by performing the converse operation of the derivative or by an integration by parts .

For others, each case must be considered as a specific case.

This table resumes antiderivatives directly determined from the inverse operation of the derivative , namey:

$$ f(x) = \int^x F'(t)dt \ \Longleftrightarrow \ F(x) = \int^x f(t)dt $$
Condition
General antiderivative
Standard functions
$$ \forall x \in \mathbb{R}, $$
$$ \int^x dt = x$$
$$ \forall x \in \mathbb{R}, $$
$$ \int^x t \ dt = \frac{x^2}{2}$$
$$ \forall x \in \hspace{0.04em} \mathbb{R}^*_+,$$
$$ \int^x \sqrt{t} \ dt = \frac{2}{3} x^{\frac{3}{2}}$$
$$ \forall x \in \hspace{0.04em} \mathbb{R}^*_+,$$
$$ \int^x \frac{1}{\sqrt{t}} \ dt = 2\sqrt{x} $$
$$ \text{when \(x\) is defined} $$
$$ \int^x t^n \ dt = \frac{x^{n+1}}{n+1}$$
$$\forall x \in \mathbb{R}, \enspace \forall n \in \mathbb{R_+^*},$$
$$ \int^x n^t \ dt = \frac{ n^x}{\ln(n)} $$
$$ \forall x \in \mathbb{R}, $$
$$ \int^x e^t \ dt = e^x$$
$$ \forall x \in \hspace{0.04em} \mathbb{R}^*, $$
$$ \int^x\frac{dt}{t} = \ln|x| $$
$$ \forall x \in \mathbb{R^*}, \enspace \forall n \in \mathbb{R^+}, $$
$$ \int^x \log_n|t| \ dt = \ln(n) \times \log_n|x| $$

\(\Longrightarrow \ \) all derivatives of standard functions

trigonometric functions
$$ \forall x \in [-1, \hspace{0.2em} 1], $$
$$ \int^x \frac{1}{\sqrt{1 - t^2}} \ dt = \operatorname{Arcsin}(x) = - \operatorname{Arccos}(x) $$
$$ \forall x \in \mathbb{R}, $$
$$ \int^x \frac{1}{\sqrt{1 + t^2}} \ dt = \operatorname{Argsinh}(x) $$
$$ \forall x \in [1, \hspace{0.1em} +\infty[, $$
$$ \int^x \frac{1}{\sqrt{t^2 - 1}} \ dt = \operatorname{Argcosh}(x) $$
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \Bigl \{ \frac{\pi}{2} + k\pi \Bigr \} \biggr] $$
$$ \int^x \Bigl[ 1 + \tan^2(t) \Bigr] \ dt = \int^x \sec^2(t) \ dt = \tan(x) $$
$$ \forall x \in \mathbb{R}, $$
$$ \int^x \frac{1}{1 + t^2} \ dt = \operatorname{Arctan}(x) $$
$$ \forall x \in \mathbb{R}, $$
$$ \int^x \operatorname{sech}^2(t) \ dt = \tanh(x) $$
$$ \forall x \in \hspace{0.04em} ]-1, \hspace{0.1em} 1[, $$
$$ \int^x \frac{1}{1 - t^2} \ dt = \operatorname{Argtanh}(x) $$
$$ \forall x \in \hspace{0.04em} ]-\infty, \hspace{0.1em} -1[\hspace{0.1em}\cup \hspace{0.1em}]1, \hspace{0.1em} +\infty[, $$
$$ \int^x \Biggl[ \frac{1}{ t^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ t^2}}} \Biggl] \ dt = -\operatorname{Arccsc}(x) $$
$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.1em} \backslash \hspace{0.2em} \left \{ 0 \right \} \Bigr] , $$
$$ \int^x \Biggl[ \frac{1}{ t^2} \times \frac{1}{ \sqrt{1 + \frac{1}{ t^2}}} \Biggl] \ dt = -\operatorname{Argcsch}(x) $$
$$ \forall x \in \hspace{0.04em} ]-\infty, \hspace{0.1em} -1[\hspace{0.1em}\cup \hspace{0.1em}]1, \hspace{0.1em} +\infty[, $$
$$ \int^x \Biggl[ \frac{1}{ t^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ t^2}}} \Biggl] \ dt = \operatorname{Arcsec}(x) $$
$$ \forall x \in \hspace{0.1em} \bigl]0, \hspace{0.1em} 1 \bigr], $$
$$ \int^x \Biggl[ \frac{1}{ t^2} \times \frac{1}{ \sqrt{\frac{1}{ t^2}- 1}} \Biggl] \ dt = -\operatorname{Argsech}(x) $$
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \bigl \{ k\pi \bigr \} \Bigr], $$
$$ \int^x \csc^2(t) \ dt = -\cot(x) $$
$$ \forall x \in \mathbb{R}, $$
$$ \int^x \frac{1}{1+t^2} \ dt = -\operatorname{Arccot}(x) $$
$$ \forall x \in \hspace{0.04em} \mathbb{R}^*, $$
$$ \int^x \Bigl[ 1 - \cot^2(t) \Bigr] \ dt = \int^x \Bigl[ -\operatorname{csch}^2(t) \Bigr] \ dt = \operatorname{coth}(x) $$
$$ \forall x \in \hspace{0.04em} ]-\infty, \hspace{0.1em} -1[ \hspace{0.1em} \cup \hspace{0.1em} ]1, \hspace{0.1em} +\infty[ , $$
$$ \int^x \frac{1}{1 - t^2} \ dt = \operatorname{Argcoth}(x) $$

\(\Longrightarrow \ \) all trigonometric derivatives

\(\Longrightarrow \ \) all trigonometric antiderivatives

operations on functions
$$ \forall a \in \mathbb{R}, \ \forall x \in \hspace{0.04em} \mathcal{D}_f, $$
$$ \int^x f(at) \ dt = \frac{1}{a} F(ax) $$
$$ \forall (\lambda, \mu) \in \hspace{0.04em} \mathbb{R}^2, \ \forall \bigl(u(x), v(x)\bigr) \in \hspace{0.04em} \mathbb{R}^2, $$
$$ \int^x \biggl[ \lambda u(t) + \mu v(t) \biggr] \ dt = \lambda U(x) + \mu V(x) $$
$$ \forall u(x) \in \hspace{0.04em} \mathbb{R}^*, $$
$$ \int^x \frac{u'(t)}{u(t)} \ dt = \ln\bigl|u(x)\bigr| $$
$$ \forall u(x) \in \hspace{0.04em} \mathbb{R}^*, $$
$$ \int^x \frac{u'(t)}{u^2(t)} \ dt = - \frac{1}{u(x)} $$
$$ \forall u(x) \in \hspace{0.04em} \mathbb{R}^*, $$
$$ \int^x \frac{u'(t)}{ 2\sqrt{u(t)}} \ dt = \sqrt{u(x)} $$

\(\Longrightarrow \ \) all derivatives of operations on functions

$$ \forall (a,b) \in D_f^2,$$
$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \Bigl[fg\Bigr]_{a}^b - \int_{a}^b (fg') \hspace{0.2em}dt $$

Let \(f : x \longmapsto f(x)\) be a \(\mathcal{C}^1\) class function on an interval \(I = \bigl[a,b \bigr]\).

Similarly, let \(\phi : t \longmapsto \phi(t) \) be a \(\mathcal{C}^1\) class function on an interval \( J \). For the change of variables to be valid, the function \(\phi\) must be a bijection from \( J \) to \( I \), thereby guaranteeing the unique existence of its inverse bounds.

$$ \forall (a,b) \in D_f^2, \ $$
$$ \int_{a}^b \hspace{0.2em} \Bigl(f \circ \phi(t)\Bigr) \hspace{0.2em} \phi'(t) \ dt = \int_{\phi(a)}^{\phi(b)} f(u) \hspace{0.2em}du $$
$$ \text{with } \begin{cases} u = \phi(t) \\ du = \phi '(t) \hspace{0.2em} dt \end{cases} $$

The integration by substitution is the transposition of the derivative of a composite function , but applied to integral calculus .

Proofs

General integration methods

Integration by parts

With the derivative of a product , we do have:

$$ \left ( f g\right)' = f'g + g'f $$
$$ f'g = ( f g)' - g'f $$

Now, thanks to the property of linearity of the integral ,

$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \int_{a}^b (fg) -\int_{a}^b (fg') \hspace{0.2em}dt $$

And,

$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \Bigl[fg\Bigr]_{a}^b -\int_{a}^b (fg') \hspace{0.2em}dt $$

Then finally,

$$ \forall (a,b) \in I^2, \enspace a < b, $$
$$ \int_{a}^b (f'g) \hspace{0.2em}dt = \Bigl[fg\Bigr]_{a}^b - \int_{a}^b (fg') \hspace{0.2em}dt $$

Integration by substitution

Let \(f : x \longmapsto f(x)\) be a \(\mathcal{C}^1\) class function on an interval \(I = \bigl[a,b \bigr]\).

Similarly, let \(\phi : t \longmapsto \phi(t) \) be a \(\mathcal{C}^1\) class function on an interval \( J \). For the change of variables to be valid, the function \(\phi\) must be a bijection from \( J \) to \( I \), thereby guaranteeing the unique existence of its inverse bounds.

Considering the following integral \( I(x) \):

$$ I(x) = \int_{a}^b f(x) \hspace{0.2em}dx = F(b) - F(a) \qquad(I(x)) $$

By performing the substitution \( x = \phi(t) \) in \( (I(x)) \), we have:

$$ x = \phi(t) \Longrightarrow \begin{cases} x \ \longrightarrow \ \phi(t) \\ dx \ \longrightarrow \ d \Bigl[ \phi(t)\Bigr] = \phi'(t) \ dt \end{cases} $$

Now, since \(\phi\) is bijective, we can introduce \( \psi : x \longmapsto \psi(x) \), its inverse function defined from \(I\) to \(J\). We then have the strict equivalence:

$$ x = \phi(t) \Longleftrightarrow t = \psi(x)$$

Thus, as \(x\) varies from \( a \) to \(b\), the variable \(t\) uniquely varies from \( \psi(a) \) to \(\psi(b)\).

That is,

$$ I(t) = \int_{\psi(a)}^{\psi(b)} \Bigl(f \circ \phi(t)\Bigr) \phi'(t) \ dt \qquad(I(t)) $$
$$ I(t) = \Bigl[ f \circ \phi(t) \Bigr]_{\psi(a)}^{\psi(b)} $$

Since the two functions \(\phi\) and \(\psi\) are inverse functions , they cancel each other out:

$$ I(t) = F\bigl(\phi(\psi(b))\bigr) - F\bigl(\phi(\psi(a))\bigr) = F(b) - F(a) $$

The two expressions \( (I(x)) \) and \( (I(t)) \) are therefore rigorously equal and are worth: \(F(b) - F(a)\).

Thus,

$$ \int_{\psi(a)}^{\psi(b)} f\Bigl( \phi(t)\Bigr) \ \phi'(t) \ dt = \int_{a}^b f(x) \hspace{0.2em}dx \qquad (1) $$

Finally, by mapping the respective bounds on each side by the function \( \phi \), equation \( (1)\) becomes \( (1')\) :

$$ \int_{\phi(\psi(a))}^{\phi(\psi(b))} f\Bigl( \phi(t)\Bigr) \ \phi'(t) \ dt = \int_{\phi(a)}^{\phi(b)} f(x) \hspace{0.2em}dx $$

And,

$$ \int_{a}^{b} f\Bigl( \phi(t)\Bigr) \ \phi'(t) \ dt = \int_{\phi(a)}^{\phi(b)} f(x) \hspace{0.2em}dx \qquad (1') $$

For simplicity, we will use \(u \) as the variable, and finally,

$$ \forall (a,b) \in D_f^2,$$
$$ \int_{a}^b \hspace{0.2em} \Bigl(f \circ \phi(t)\Bigr) \hspace{0.2em} \phi'(t) \ dt = \int_{\phi(a)}^{\phi(b)} f(u) \hspace{0.2em}du $$
$$ \text{with } \begin{cases} u = \phi(t) \\ du = \phi '(t) \hspace{0.2em} dt \end{cases} $$

Be careful not to confuse the direct change of variables:

$$ \begin{cases} u = \phi(t) \\ du = \phi '(t) \hspace{0.2em} dt \end{cases} $$

with an indirect change of variables:

$$ \begin{cases} t = \psi(u) \\ dt = \psi '(u) \hspace{0.2em} du \end{cases} $$

General integration methods recap table

Examples

  1. Integration by parts examples

    1. Example 1
      $$ \alpha = \int_{0}^{\frac{\pi}{2}} t \cdot \sin(t) \hspace{0.2em}dt $$

      We choose to take \(f\) and \(g'\) such as:

      $$ \Biggl \{ \begin{gather*} f(t) = t \\ g'(t) = \sin(t) \end{gather*} $$
      $$ \Biggl \{ \begin{gather*} f'(t) = 1 \\ g(t) = -\cos(t) \end{gather*} $$
      $$ \alpha = \Bigl[-t.\cos(t) \Bigr]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} - \cos(t) \hspace{0.2em}dt $$
      $$ \alpha = \Bigl[-t.\cos(t) + \sin(t) \Bigr]_{0}^{\frac{\pi}{2}} $$
      $$ \alpha = - \frac{\pi}{2} \times 0 + 1 - (0\times 1 + 0 ) $$
      $$ \alpha = 1 $$
    2. Example 2
      $$ \beta = \int_{1}^{e} \ln(t) \hspace{0.2em}dt $$

      We choose to take \(f\) and \(g'\) such as:

      $$ \Biggl \{ \begin{gather*} f(t) = \ln(t) \\ g'(t) = 1 \end{gather*} $$
      $$ \Biggl \{ \begin{gather*} f'(t) = \frac{1}{t} \\ g(t) = t \end{gather*} $$
      $$ \beta = \Bigl[t.\ln(t) \Bigr]_{1}^{e} - \int_{1}^{e} \hspace{0.2em}dt $$
      $$ \beta = \Bigl[t.\ln(t) \Bigr]_{1}^{e}- \bigl[t \bigr]_{1}^{e} $$
      $$ \beta = e . \ln(e) - \ln(1) - (e -1) $$
      $$ \beta =1 $$
    3. Example 3: integrate a function of type \(f(t) e^t\)

      Notably while solving ordinary differential equations , we may determine antiderivatives of type:

      $$ \int^{x} f(t) \hspace{0.1em} e^t \hspace{0.2em}dt $$

      For instance, with a polynomial function \(f(t)\), we integrate many times in a row until the degree decreases to reach \(0\).

      We choose to integrate the exponential function so that the polynomial function is the one which is derivated and decrease in degree as integrations progresses.

      Let us integrate this polynomial-exponential function:

      $$ \gamma = \int^{x} (4t^3 - t + 4) \hspace{0.1em} e^{2t} \hspace{0.2em}dt $$
      $$ \Biggl \{ \begin{gather*} f(t) = 4t^3 - t +4 \\ g'(t) = e^{2t} \end{gather*} $$
      $$ \Biggl \{ \begin{gather*} f'(t) = (12t^2 -1) \\ g(t) = \frac{e^{2t}}{2} \end{gather*} $$
      $$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - \int^{x} ( 12t^2 -1 ) \hspace{0.1em} \frac{e^{2t}}{2} \hspace{0.2em}dt $$

      We take this oppotunity to take the constant out of it.

      $$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 6\int^{x} t^2 \hspace{0.1em} e^{2t} \hspace{0.2em}dt + \int^{x} \frac{e^{2t}}{2} \hspace{0.2em}dt $$
      $$ \Biggl \{ \begin{gather*} f(t) = t^2 \\ g'(t) = e^{2t} \end{gather*} $$
      $$ \Biggl \{ \begin{gather*} f'(t) = 2t \\ g(t) = \frac{e^{2t}}{2} \end{gather*} $$
      $$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 6 \Biggl( \Biggl[ t^2 \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - \int^{x} t \hspace{0.1em} e^{2t} \hspace{0.2em}dt \Biggr) + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$
      $$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 3 \Biggl[ t^2 \hspace{0.1em} e^{2t} \Biggr]^{x} + 6 \int^{x} t \hspace{0.1em} e^{2t} \hspace{0.2em}dt + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$
      $$ \Biggl \{ \begin{gather*} f(t) = t \\ g'(t) = e^{2t} \end{gather*} $$
      $$ \Biggl \{ \begin{gather*} f'(t) = 1 \\ g(t) = \frac{e^{2t}}{2} \end{gather*} $$
      $$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 3 \Biggl[ t^2 \hspace{0.1em} e^{2t}\Biggr]^{x} + 6\Biggl[t \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 6 \int^{x} \frac{e^{2t}}{2} \hspace{0.2em}dt + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$
      $$ \gamma = \Biggl[(4t^3 - t +4 ) \hspace{0.1em} \frac{e^{2t}}{2} \Biggr]^{x} - 3 \Biggl[ t^2 \hspace{0.1em} e^{2t}\Biggr]^{x} + 3\Biggl[t \hspace{0.1em} e^{2t} \Biggr]^{x} - 3\Biggl[ \frac{e^{2t}}{2} \Biggr]^{x} + \Biggl[ \frac{e^{2t}}{4} \Biggr]^{x} $$
      $$ \gamma = \frac{1}{2} \left (4x^3 - x +4 \right) \hspace{0.1em} e^{2x} - 3x^2 e^{2x} + 3x e^{2x} - \frac{5}{4}e^{2x} $$

      We finally factorize it all.

      $$ \gamma = e^{2x} \left (2x^3 - \frac{1}{2}x +2 - 3x^2 + 3x - \frac{5}{4} \right) \hspace{0.1em} $$
      $$ \gamma = e^{2x} \left (2x^3 -3x^2 + \frac{5}{2} x + \frac{3}{4} \right) \hspace{0.1em} $$

  2. Integration by substitution examples

    1. Example 1
      $$ A = \int_{0}^{\frac{\pi}{2}} \frac{\sin(t)}{1 + 3 \cos^2(t)} \ dt$$

      In the case of a trigonometric polynomial function, we can apply Bioche's rules .

      In our case, the integrand \( \omega(t) \) is invariant by the variable change \( t = -t \):

      $$ \omega(-t) = \frac{\sin(-t)}{1 + 3 \cos^2(-t)} \ d(-t) $$
      $$ \omega(-t) = \frac{-\sin(t)}{1 + 3 \cos^2(t)} \times (-dt) $$
      $$ \omega(-t) = \frac{\sin(t)}{1 + 3 \cos^2(t)} \ dt $$
      $$ \omega(-t) = \omega(t) $$

      We thus set a new variable down:

      $$ u = \cos(t)$$

      We do have now:

      $$ \Biggl \{ \begin{gather*} u = \cos(t) \\ du = -\sin(t) \ dt \end{gather*} $$

      We replace it in our main expression, without forgetting the bounds:

      $$ A = \int_{\cos(0)}^{\cos(\frac{\pi}{2})} \frac{-du}{1 + 3 u^2} $$
      $$ A = -\int_{1}^{0} \frac{1}{1 + 3 u^2} \ du $$

      Thanks to this property of the integrals , we know that:

      $$ \forall (a,b) \in I^2, \ \int_{b}^a f(t) \hspace{0.2em}dt = -\int_{a}^b f(t) \hspace{0.2em}dt $$
      $$ A = \int_{0}^{1} \frac{1}{1 + 3 u^2} \ du$$

      The antiderivative of the \(\operatorname{Arctan}\) function is known as:

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

      So, in our case,

      $$ A = \frac{1}{3} \int_{0}^{1} \frac{1}{\frac{1}{3} + u^2} \ du$$

      Then,

      $$ A = \frac{1}{3} \int_{0}^{1} \frac{1}{ \left(\frac{1}{\sqrt{3}}\right)^2 + \ u^2} \ du = \frac{1}{3} \Bigl[\sqrt{3} \ \operatorname{Arctan}\left(\sqrt{3} u\right) \Bigr]_0^1 $$
      $$ A = \frac{1}{3} \ \Bigl[ \sqrt{3} \ \operatorname{Arctan}(\sqrt{3}) - \sqrt{3} \ \operatorname{Arctan}(0) \Bigr] $$
      $$ A = \frac{\sqrt{3}}{3} \operatorname{Arctan}(\sqrt{3}) $$
      $$ A = \frac{\sqrt{3}\pi}{9} $$
    2. Example 2
      $$ B = \int_{0}^{1} \ \frac{t^2}{1 + t^3} \ dt$$

      The idea is to set \( u = t^3 \) down to get \( t^2 \ dt\) at the numerator (mutiplied by a constant) after substitution.

      $$ \Biggl \{ \begin{gather*} u = t^3 \\ du = 3t^2 \ dt \end{gather*} $$

      We replace it all (bounds do not change because \( 0\) and \( 1 \) do not vary taking their cube):

      $$ B = \int_{0}^{1} \ \frac{1}{1 + u} \ \frac{du}{3} $$
      $$ B = \frac{1}{3} \int_{0}^{1} \ \frac{1}{1 + u} \ du$$
      $$ B = \frac{1}{3} \Bigl[\ln |1 + u| \Bigr]_{0}^{\ 1} $$
      $$ B = \frac{1}{3} \ln(2) $$
    3. Example 3
      $$ C = \int_{1}^{3} \ \frac{e^{2t}}{1 - e^t} \ dt$$

      The integrand in only defined for \( \mathbb{R} \ \backslash \ \{0 \}\).

      $$ \Biggl \{ \begin{gather*} u = e^t \\ du = e^t \ dt \end{gather*} $$
      $$ C = \int_{e}^{e^3} \ \frac{u}{1 - u} \ du$$

      Let us use a little trick to transform the expression:

      $$ C = \int_{e}^{e^3} \ \frac{u + 1 - 1}{1 - u} \ du$$
      $$ C = \int_{e}^{e^3} \ \frac{u - 1}{1 - u} + \frac{1}{1 - u} \ du$$
      $$ C = \int_{e}^{e^3} \ \frac{-(1-u)}{1 - u} + \frac{1}{1 - u} \ du$$
      $$ C = -\int_{e}^{e^3} du + \int_{e}^{e^3} \frac{1}{1 - u} \ du$$
      $$ C = - \Bigl[u\Bigr]_{e}^{e^3} + \Bigl[ - \ln |1 - u| \Bigr]_{e}^{e^3} $$
      $$ C = - e^3 + e - \ln |1 - e^3| + \ln |1 - e| $$
      $$ C = - e^3 + e - \ln (e^3-1) + \ln(e-1) $$
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