$\int \mathrm{e}^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} \mathrm{~d} x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} I &=\int e^{x}\left(\frac{1-x}{1+x^{2}}ight)^{2} d x \ &=\int e^{x} \frac{\left(1+x^{2}-2 xight)}{\left(1+x^{2}ight)^{2}} d x=\int e^{x}\left[\frac{1+x^{2}}{\left(1+x^{2}ight)^{2}}-\frac{2 x}{\left(1+x^{2}ight)^{2}}ight] d x \ &=\int e^{x}\left[\frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}ight)^{2}}ight] d x=\frac{e^{x}}{1+x^{2}}+C \end{aligned}$

$\int \mathrm{e}^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} \mathrm{~d} x=$

Solution