# How do you integrate ∫(x^3ex^2)/(x^2 +1)^2dx?

## ∫(x^3e^(x^2))/(x^2 +1)^2dx  ?

Jul 12, 2018

$\int \frac{{x}^{3} {e}^{{x}^{2}}}{{x}^{2} + 1} ^ 2 \mathrm{dx} = {e}^{{x}^{2}} / \left(2 \left({x}^{2} + 1\right)\right) + c$

#### Explanation:

Here,

$I = \int \frac{{x}^{3} {e}^{{x}^{2}}}{{x}^{2} + 1} ^ 2 \mathrm{dx}$

$= \int \frac{{x}^{2} {e}^{{x}^{2}}}{{x}^{2} + 1} ^ 2 \cdot x \mathrm{dx}$

Subst . color(violet)(x^2=u=>2xdx=du=>xdx=1/2du

So,

$I = \int \frac{u {e}^{u}}{u + 1} ^ 2 \cdot \frac{1}{2} \mathrm{du}$

$= \frac{1}{2} \int \frac{\left(u + 1 - 1\right) {e}^{u}}{u + 1} ^ 2 \mathrm{du}$

$= \frac{1}{2} \int \left\{\frac{\left(u + 1\right)}{u + 1} ^ 2 - \frac{1}{u + 1} ^ 2\right\} {e}^{u} \mathrm{du}$

=1/2{color(blue)(int1/(u+1)e^udu)-color(red) (int1/(u+1)^2e^udu)}

Using Integration by parts: in the first integral

$I$=1/2{color(blue)([1/(u+1)inte^u-int(-1)/(u+1)^2e^udu])- color(red)(int1/(u+1)^2e^udu)}

$I$=1/2{color(brown)(1/(u+1)e^u+color(blue) (int(1)/(u+1)^2e^udu)-color(red)(int1/(u+1)^2e^udu)}

$I = \frac{1}{2} \cdot \textcolor{b r o w n}{\frac{1}{u + 1} {e}^{u}} + c$

Subst. back color(violet)( u=x^2 we get

$I = \frac{1}{2} \left(\frac{1}{{x}^{2} + 1}\right) {e}^{{x}^{2}} + c$

Hence ,

$I = {e}^{{x}^{2}} / \left(2 \left({x}^{2} + 1\right)\right) + c$
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Note: Change in the question is as below.

${x}^{3} e {x}^{2} \to {x}^{3} {e}^{{x}^{2}}$

Jul 12, 2018

The answer is $= \frac{{e}^{{x}^{2}}}{2 \left({x}^{2} + 1\right)} + C$

#### Explanation:

Perform the substitution

$u = {x}^{2}$, $\implies$, $\mathrm{du} = 2 x \mathrm{dx}$

The integral is

$I = \int \frac{{x}^{3} {e}^{{x}^{2}} \mathrm{dx}}{{x}^{2} + 1} ^ 2$

$= \frac{1}{2} \int \frac{u {e}^{u} \mathrm{du}}{u + 1} ^ 2$

Perform an integration by parts

$\int w v ' \mathrm{dx} = w v - \int w ' v \mathrm{dx}$

$w = u {e}^{u}$, $\implies$, $w ' = u {e}^{u} + {e}^{u} = {e}^{u} \left(u + 1\right)$

$v ' = \frac{1}{u + 1} ^ 2$, $\implies$, $v = - \frac{1}{u + 1}$

Therefore,

$I = - \frac{1}{2} \frac{u {e}^{u}}{u + 1} + \frac{1}{2} \int {e}^{u} \mathrm{du}$

$= {e}^{u} / 2 - \frac{u {e}^{u}}{2 \left(u + 1\right)}$

$= {e}^{{x}^{2}} / 2 - \frac{{x}^{2} {e}^{{x}^{2}}}{2 \left({x}^{2} + 1\right)} + C$

$= \frac{1}{2} {e}^{{x}^{2}} \frac{\left({x}^{2} + 1 - {x}^{2}\right)}{{x}^{2} + 1} + C$

$= \frac{{e}^{{x}^{2}}}{2 \left({x}^{2} + 1\right)} + C$