Integrated and Frustrated

Solution:

$\int_0^1 \frac{1-e^{-x}}{1+e^{-x}}dx=\int_0^1\frac{1}{1+e^{-x}}dx-\int_0^1\frac{e^{-x}}{1+e^{-x}}dx=\int_0^1\frac{e^x}{e^x+1}dx-\int_0^1\frac{e^{-x}}{1+e^{-x}}dx$ $u=1+e^x \to du=e^xdx$ $v=1+e^{-x} \to dv=-e^{-x}dx$ $\int_{x=0}^{x=1} \frac{1}{u}du+\int_{x=0}^{x=1} \frac{1}{v}dv=\ln (1+e^x)|_0^1+\ln (1+e^{-x})|_0^1=\ln[(1+e^x)(1+e^{-x})]|_0^1=$ $\ln [(1+e^1)(1+e^{-1})]-\ln [(1+e^0)(1+e^{-0})]=\ln\left[\frac{(1+e)(1+\frac{1}{e})}{4}\right] \approx \boxed{0.24}$