## Calculus: Early Transcendentals (2nd Edition)

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