The correct option is A e^x[1x+1]+c I = ∫xe^x/(x+1)^2dx =∫(x+1 1)e^x/(x+1)^2dx =∫(x+1)e^x/(x+1)^2dx ∫e^x/(x+1)^2dx =∫e^x/x+1 dx ∫e^x/ ...

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Integration by Parts

Evaluate ∫x...

Question

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

e^{x} [\frac{1}{x + 1}] + c

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e^{x} [\frac{- 1}{x + 1}] + c

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e^{x} x + c

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

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Solution

The correct option is A $e^{x} [\frac{1}{x + 1}] + c$
$I = \int \frac{x e^{x}}{(x + 1)^{2}} d x$
$= \int \frac{(x + 1 - 1) e^{x}}{(x + 1)^{2}} d x$
$= \int \frac{(x + 1) e^{x}}{(x + 1)^{2}} d x - \int \frac{e^{x}}{(x + 1)^{2}} d x$
$= \int \frac{e^{x}}{x + 1} d x - \int \frac{e^{x}}{(x + 1)^{2}} d x$

We know, $\int u . v d x = u \int v d x$ $- \int (\frac{d u}{d x} \int v d x) d x$

Now by integrating the functions by parts, for the first part taking $e^{x}$ as $u$ and $\frac{1}{x + 1}$ as $v$

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

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