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

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

$\displaystyle \int e^x\left(\frac{1-x}{1+x^2}\right)^2 dx.$

$\text{Consider } \frac{d}{dx}\left(\frac{e^x}{1+x^2}\right).$

$=\frac{e^x(1+x^2)-e^x(2x)}{(1+x^2)^2}.$

$=\frac{e^x(1-2x+x^2)}{(1+x^2)^2} = e^x\left(\frac{1-x}{1+x^2}\right)^2.$

$\text{ therefore } \int e^x\left(\frac{1-x}{1+x^2}\right)^2 dx =\frac{e^x}{1+x^2}+C.$

$\displaystyle \int e^x\left(\frac{1-x}{1+x^2}\right)^2 dx =\frac{e^x}{1+x^2}+C.$