Question

$\underset{0}{\overset{1}{\int }}\frac{e^x}{1+e^{2x}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}{e}^x=t.\mathrm{Then},e^{x}dx=dt \\ \mathrm{When}x=0,t=1\mathrm{and}x=1,t=e \\ \therefore I={\int }_0^1\frac{e^x}{1+e^{2x}}\mathit{}\mathit{d}x \\ \Rightarrow I={\int }_1^e\frac{dt}{1+t^2} \\ \Rightarrow I={\left[{\tan }^{-1}x\right]}_1^e \\ \Rightarrow I={\tan }^{-1}e-{\tan }^{-1}1 \\ \Rightarrow I={\tan }^{-1}e-\frac{\mathrm{\pi }}{4} \end{gathered}$$