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

MCQ

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

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

✓

Answer

Correct option: A.

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

(A) $\frac{e^x}{1+x^2}+C$
$\begin{aligned} \int e^x\left(\frac{1-x}{1+x^2}\right)^2 d x & =\int e^x \frac{1+x^2-2 x}{\left(1+x^2\right)^2} d x \\ & =\int e^x\left[\frac{1}{\left(1+x^2\right)}-\frac{2 x}{\left(1+x^2\right)^2}\right] d x \\ & =\int e^x\left[f(x)+f^{\prime}(x)\right] d x,\end{aligned}$
[where $f(x)=\frac{1}{1+x^2}$ ]
$\begin{array}{l}=e^x f(x)+C \\ =\frac{e^x}{1+x^2}+C\end{array}$

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