If $\int\limits_0^1\frac{e^x}{1+x}dx=m$, then the value of $\int\limits_0^1\frac{e^x}{(1+x)^2}dx$ is

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

The correct answer is Option (3) → $m-\frac{e}{2}+1$

Given $\displaystyle \int_{0}^{1}\frac{e^{x}}{1+x}\,dx=m$

We need $\displaystyle \int_{0}^{1}\frac{e^{x}}{(1+x)^{2}}\,dx$

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

Integrate by parts: let $u=e^{x}$, $dv=\frac{dx}{(1+x)^{2}}$

$\Rightarrow du=e^{x}dx,\ v=-\frac{1}{1+x}$

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

$I=\left(-\frac{e}{2}+1\right)+m$

$I=m+1-\frac{e}{2}$