e^x (1-x 1+x^2 )^2 d x is equal to

MCQ

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

A
$-\frac{e^x}{1+x^2}+C$
B
$-\frac{e^x}{\left(1+x^2\right)^2}+C$
✓
$\frac{e^x}{1+x^2}+C$
D
$\frac{e^x}{\left(1+x^2\right)^2}+C$

✓

Answer

Correct option: C.

$\frac{e^x}{1+x^2}+C$

(c) $\frac{e^x}{1+z^2}+C$
Explanation: Given $\int e^x\left(\frac{1-x}{1+x^2}\right)^2 d x$
$
\begin{array}{l}
\Rightarrow \int e^x\left(\frac{1-x}{1+x^2}\right)^2 d x=\int e^x\left(\frac{1+x^2-2 z}{\left(1+x^2\right)^2}\right) d x \\
\Rightarrow \int e^x\left(\frac{1+x^2-2 x}{\left(1+x^2\right)^2}\right) d x=\int e^x\left\{\left(\frac{1+x^2}{\left(1+x^2\right)^2}\right)+\left(\frac{-2 x}{\left(1+x^2\right)^2}\right)\right\} d x \\
=\int e^x\left\{\left(\frac{1}{\left(1+x^2\right)}\right)+\left(\frac{-2 x}{\left(1+x^2\right)^2}\right)\right\} d x
\end{array}
$
Now using the property: $\int e^x\left(f(x)+f^{\prime}(x)\right) d x=e^x f(x)$
Now in $\int e^x\left\{\left(\frac{1}{\left(1+x^2\right)}\right)+\left(\frac{-2 x}{\left(1+x^2\right)^2}\right)\right\} d x$
$
\begin{array}{l}
\Rightarrow f(x)=\frac{1}{\left(1+x^2\right)} \\
\Rightarrow f^{\prime}(x)=\frac{-2 x}{\left(1+x^2\right)^2} \\
\Rightarrow \int e^{x}\left\{\left(\frac{1}{\left(1+x^2\right)}\right)+\left(\frac{-2 x}{\left(1+x^2\right)^2}\right)\right\} dx=\frac{e^{x}}{1+x^2}+C \\
\Rightarrow \int e^x\left(\frac{1-x}{1+x^2}\right)^2 d x=\frac{e^2}{1+x^2}+C
\end{array}
$

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