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

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

The correct answer is option (1) → $\frac{e^x}{1+x^2}+C$

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

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

$f(x)=\frac{1}{1+x^2},f'(x)=\frac{2x}{(1+x^2)^2}$

so $I=e^xf(x)+C=\frac{e^x}{1+x^2}+C$