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>