The value of the definite integral $\int\limits_0^1e^x\frac{(1-x)^2}{(1+x^2)^2}dx$ is:

$\frac{e}{2}-1$

The correct answer is Option (2) → $\frac{e}{2}-1$

Given

$I=\displaystyle\int_{0}^{1} e^{x}\,\frac{(1-x)^{2}}{(1+x^{2})^{2}}\,dx$

Let $g(x)=\frac{1}{1+x^{2}}$. Then

$g'(x)=-\frac{2x}{(1+x^{2})^{2}}$

and

$g(x)+g'(x)=\frac{1}{1+x^{2}}-\frac{2x}{(1+x^{2})^{2}}=\frac{(1-x)^{2}}{(1+x^{2})^{2}}$

Therefore

$e^{x}\frac{(1-x)^{2}}{(1+x^{2})^{2}}=e^{x}\big(g(x)+g'(x)\big)=\frac{d}{dx}\!\left(e^{x}g(x)\right)$

So

$I=\left[e^{x}\frac{1}{1+x^{2}}\right]_{0}^{1}=\frac{e}{2}-1$