∫xex(1+x)2dx is equal to

# $\int \frac{{\mathrm{xe}}^{\mathrm{x}}}{\left(1+\mathrm{x}{\right)}^{2}}\mathrm{dx}$ is equal to

1. A

$\frac{{\mathrm{e}}^{\mathrm{x}}}{1+\mathrm{x}}+\mathrm{C}$

2. B

$\frac{-1}{1+\mathrm{x}}+\mathrm{C}$

3. C

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

4. D

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

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### Solution:

Let $\begin{array}{r}\mathrm{I}=\int \frac{{\mathrm{xe}}^{\mathrm{x}}}{\left(1+\mathrm{x}{\right)}^{2}}\mathrm{dx}=\int {\mathrm{e}}^{\mathrm{x}}\frac{\left(\mathrm{x}+1\right)-1}{\left(1+\mathrm{x}{\right)}^{2}}\mathrm{dx}\\ =\int {\mathrm{e}}^{\mathrm{x}}\left[\frac{1}{\left(1+\mathrm{x}\right)}-\frac{1}{\left(1+\mathrm{x}{\right)}^{2}}\right]\mathrm{dx}\end{array}$

Let $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{1+\mathrm{x}}⇒{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=-\frac{1}{\left(1+\mathrm{x}{\right)}^{2}}$

we know that $\int {\mathrm{e}}^{\mathrm{x}}\left\{\mathrm{f}\left(\mathrm{x}\right)+{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\right\}\mathrm{dx}={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)$

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