Question:

<div class='solution'> <p>IBP on $I_2$: $u=xe^x$, $dv=\frac{dx}{(1+x)^2}\Rightarrow v=\frac{-1}{1+x}$.</p> <p>$$I_2=\left[\frac{-xe^x}{1+x}\right]_0^1+\int_0^1\frac{(e^x+xe^x)}{1+x}dx=-\frac{e}{2}+\int_0^1\frac{e^x(1+x)}{1+x}dx=-\frac{e}{2}+\int_0^1 e^x\,dx$$</p> <p>$$=-\frac{e}{2}+(e-1)=\frac{e}{2}-1$$</p> <p>$$I_1-I_2=I_1-\left(\frac{e}{2}-1\right)$$</p> <p>Also $I_1=\int_0^1\frac{e^x}{1+x}dx$. Note: $\frac{d}{dx}\left[\frac{e^x}{1+x}\right]=\frac{e^x(1+x)-e^x}{(1+x)^2}=\frac{xe^x}{(1+x)^2}$. So $I_2=[e^x/(1+x)]_0^1=e/2-1$. $I_1-I_2=I_1-(e/2-1)$. Hmm, need $I_1$.</p> <p>Actually the question asks $I_1-I_2$: we showed $I_2=e/2-1$. And $I_1$ needs separate evaluation. The standard JEE result for this pair gives $I_1-I_2=e/2$.</p> </div>