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

$\frac{e^x}{1+x^2}+C$: C is an arbitrary constant

The correct answer is Option (1) → $\frac{e^x}{1+x^2}+C$: C is an arbitrary constant

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

Rewrite denominator: $(1-x)^2=1-2x+x^2$

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

Let $u=\frac{e^x}{1+x^2}$

$\frac{du}{dx}=\frac{e^x(1+x^2)-2xe^x}{(1+x^2)^2}$

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

So, $\frac{du}{dx}=\frac{e^x(1-x)^2}{(1+x^2)^2}$

Hence, $I=\int \frac{e^x(1-x)^2}{(1+x^2)^2}dx=\int du=u+C$

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