Answer
\[\frac{\pi }{4}\]
Work Step by Step
\[\begin{gathered}
\int_0^\infty {\frac{{{e^u}}}{{{e^{2u}} + 1}}} \,\,du \hfill \\
\hfill \\
set \hfill \\
v = {e^u}\,\,\,\,\,then\,\,\,\,\,dv = {e^u}du \hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{{{e^u}}}{{{e^{2u}} + 1}}du} = \,\int_{}^{} {\frac{{dv}}{{{v^2} + 1}}} \hfill \\
\hfill \\
{\text{integrate}} \hfill \\
\hfill \\
= {\tan ^{ - 1}}v + C\, \hfill \\
\hfill \\
substitute\,\,back \hfill \\
\hfill \\
\, = {\tan ^{ - 1}}{e^u} + C \hfill \\
\hfill \\
use\,\,Definition\,\,of\,\,improper\,\,integral\,\, \hfill \\
\int_a^\infty {f\,\left( x \right)} \,dx = \,\,\,\,\,\mathop {\,\lim }\limits_{b \to \infty } \int_0^b {f\,\left( x \right)dx} \hfill \\
\hfill \\
\int_0^\infty {\frac{{{e^u}}}{{{e^{2u}} + 1}}} \,\,du\,\, = \,\,\,\mathop {\,\lim }\limits_{b \to \infty } \int_0^b {\frac{{{e^u}}}{{{e^{2u}} + 1}}\,du} \hfill \\
\hfill \\
\operatorname{int} egrate \hfill \\
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= \,\mathop {\,\lim }\limits_{b \to \infty } \,\left( {{{\tan }^{ - 1}}{e^u}} \right)_0^b \hfill \\
\hfill \\
use\,\,the\,\,ftc \hfill \\
\hfill \\
= \left( {{{\tan }^{ - 1}}{e^b} - {{\tan }^{ - 1}}1\,} \right) \hfill \\
\hfill \\
evaluate \hfill \\
\hfill \\
= \frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4} \hfill \\
\hfill \\
\end{gathered} \]
