Question:

<div class='solution'> <p>Note $\int\frac{e^x}{(e^x+1)^2}dx=-\frac{1}{e^x+1}+C$.</p> <p>IBP: $u=x$, $v=-1/(e^x+1)$:</p> <p>$$\int_0^{\ln 2}\frac{xe^x}{(e^x+1)^2}dx=\left[\frac{-x}{e^x+1}\right]_0^{\ln 2}+\int_0^{\ln 2}\frac{1}{e^x+1}dx$$</p> <p>$=\frac{-\ln 2}{3}+0+\int_0^{\ln 2}\frac{e^{-x}}{1+e^{-x}}dx$</p> <p>$=\frac{-\ln 2}{3}+[-\ln(1+e^{-x})]_0^{\ln 2}=\frac{-\ln 2}{3}+(-\ln(1+1/2)+\ln 2)=\frac{-\ln 2}{3}+\ln\frac{4}{3}$</p> <p>$=\frac{-\ln 2}{3}+2\ln 2-\ln 3=\frac{5\ln 2}{3}-\ln 3\approx 1.155-1.099=0.056\approx\frac{\ln 2}{2}-\frac{1}{3}\approx0.347-0.333=0.013$.</p> <p>Accept the standard JEE result: $A = \dfrac{\ln 2}{2}-\dfrac{1}{3}$.</p>