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If $\int \frac{x e^{x}}{(1+x)^{2}} d x=e^{x} f(x)+c$, then $f(x)$ is equal to
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1243 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{1+x}$
We have, $I=\int \frac{x e^{x}}{(1+x)^{2}} d x$
$$
=\int \frac{(x+1-1)}{(x+1)^{2}} e^{x} d x
$$
$$
\begin{aligned}
&=\int\left[\frac{1}{x+1}-\frac{1}{(x+1)^{2}}\right] e^{x} d x \\
&=e^{x} \frac{1}{(1+x)}+C
\end{aligned}
$$
On comparing, we get
$$
f(x)=\frac{1}{1+x}
$$
$$
=\int \frac{(x+1-1)}{(x+1)^{2}} e^{x} d x
$$
$$
\begin{aligned}
&=\int\left[\frac{1}{x+1}-\frac{1}{(x+1)^{2}}\right] e^{x} d x \\
&=e^{x} \frac{1}{(1+x)}+C
\end{aligned}
$$
On comparing, we get
$$
f(x)=\frac{1}{1+x}
$$
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