∫ e^x \((\frac{1-x}{1+x^2})^2\)dx =

\(\frac{e^x}{1+x^2}\)+C

$I=∫ e^x(\frac{(1-x)}{(1+x^2)^2})dx$

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

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

It's in the form $∫e^x(f(x)+f'(x))dx=e^xf(x)+c$

where $f(x)=\frac{1}{1+x^2}$ such that $f'(x)=\frac{-2x}{(1+x^2)^2}$

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

So option 1 is correct.