$\int \frac{x e^{x}}{(1 + x)^{2}} d x$ is equal to

\frac{- e^{x}}{x + 1} + C

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

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

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\frac{- x e^{x}}{x + 1} + C

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\frac{e^{x}}{(x + 1)^{2}} + C

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Solution

The correct option is E $\frac{e^{x}}{x + 1} + C$
Let $I = \int \frac{x e^{x}}{(1 + x)^{2}} d x$

$I = \int \frac{(x + 1 - 1) e^{x}}{(1 + x)^{2}} d x$

$I = \int \frac{e^{x}}{(1 + x)} d x - \int \frac{e^{x}}{(1 + x)^{2}} d x$
Applying integration by parts in first integral, we get
$I = \frac{e^{x}}{(1 + x)} - \int - (\frac{1}{(1 + x)^{2}}) e^{x} d x - \int \frac{e^{x}}{(1 + x)^{2}} d x + C$

$I = \frac{e^{x}}{(1 + x)} + \int (\frac{1}{(1 + x)^{2}}) e^{x} d x - \int \frac{e^{x}}{(1 + x)^{2}} d x + C$
$I = \frac{e^{x}}{(1 + x)} + C$

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