I=∫xex1+x2dx=∫exx+1−1(x+1)2dx∵∫exf(x)+f′(x)dx=exf(x)+cI=∫ex1x+1−1(x+1)2dx=ex1x+1+c=exx+1+c

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

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$e^{x} + c$

$\frac{- e^{x}}{{(x + 1)}^{2}} + c$

$\frac{e^{x}}{(x + 1)} + c$

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

answer is D.

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Detailed Solution

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

$\begin{array}{l} = \int e^{x} \{\frac{x + 1 - 1}{(x + 1)^{2}}\} d x \\ (∵ \int e^{x} \{f (x) + f^{'} (x)\} d x = e^{x} f (x) + c) \\ I = \int e^{x} (\frac{1}{x + 1} - \frac{1}{(x + 1)^{2}}) d x \\ = e^{x} (\frac{1}{x + 1}) + c \\ = \frac{e^{x}}{x + 1} + c \end{array}$