Answer
$$
\int_{0}^{1} \frac{e^{x}}{1+e^{2 x}} d x=\arctan e-\frac{\pi}{4}
$$
Work Step by Step
$$
\int_{0}^{1} \frac{e^{x}}{1+e^{2 x}} d x
$$
$$\text { Let } u=e^{x} . \text { Then } d u=e^{x} d x \\
x : 0 \rightarrow 1 . \text { Then } u : 1 \rightarrow e $$
substituting in the given integral we have :
$$
\begin{aligned}
\int_{0}^{1} \frac{e^{x}}{1+e^{2 x}} d x &=\int_{1}^{e} \frac{1}{1+u^{2}} d u \\
&=[\arctan u]_{1}^{e} \\
& =\arctan e-\arctan 1\\
&=\arctan e-\frac{\pi}{4}
\end{aligned}
$$
