What is $\int \frac{x e^{x} d x}{(x + 1)^{2}}$ equal to?
where $c$ is the constant of integration.

(x + 1)^{2} e^{x} + c

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(x + 1) e^{x} + c

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\frac{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 B $\frac{e^{x}}{x + 1} + c$
To find $\int \frac{x e^{x} d x}{{(x + 1)}^{2}}$
Let $\frac{e^{x}}{x + 1} = t$ $\Rightarrow \frac{x e^{x} d x}{{(x + 1)}^{2}} = d t$
$\Rightarrow \int \frac{x e^{x} d x}{{(x + 1)}^{2}} = \int d t = t + c = \frac{e^{x}}{x + 1} + c$

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