## Calculus: Early Transcendentals (2nd Edition)

$= \frac{{3{\pi ^2}}}{{32}}$
$\begin{gathered} \int_1^\infty {\frac{{{{\tan }^{ - 1}}s}}{{{s^2} + 1}}\,ds} \hfill \\ \hfill \\ finding\,\,the\,\,anti\,derivative \hfill \\ \hfill \\ u = {\tan ^{ - 1}}s\,\,\,\,then\,\,\,\,du = \frac{1}{{{s^2} + 1}}\,ds \hfill \\ \hfill \\ \int_{}^{} {\frac{{{{\tan }^{ - 1}}s}}{{{s^2} + 1}}\,ds} \,\, = \int_{}^{} {udu} \hfill \\ \hfill \\ \operatorname{integrate} \,\, \hfill \\ \hfill \\ = \frac{{{u^2}}}{2} + C\,\, \hfill \\ \hfill \\ substitute\,\,back \hfill \\ \hfill \\ = \frac{1}{2}\,{\left( {{{\tan }^{ - 1}}s} \right)^2} + C \hfill \\ \hfill \\ use\,\,the\,\,Definition\,\,of\,\,improper\,\,integral\, \hfill \\ \hfill \\ \int_a^\infty {f\,\left( x \right)} \,dx = \,\,\,\,\,\mathop {\,\lim }\limits_{b \to \infty } \int_a^b {f\,\left( x \right)dx} \hfill \\ \hfill \\ \int_1^\infty {\frac{{{{\tan }^{ - 1}}s}}{{{s^2} + 1}}\,ds} \,\, = \,\,\,\,\,\mathop {\,\lim }\limits_{b \to \infty } \,\,\int_1^b {\frac{{{{\tan }^{ - 1}}s}}{{{s^2} + 1}}\,ds} \hfill \\ \hfill \\ = \,\,\,\mathop {\,\lim }\limits_{b \to \infty } \,\,\,\left[ {\frac{1}{2}\,{{\left( {{{\tan }^{ - 1}}\,\,s} \right)}^2}} \right]_2^b \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = \,\,\,\mathop {\,\lim }\limits_{b \to \infty } \,\,\,\left[ {\frac{1}{2}\,{{\left( {{{\tan }^{ - 1}}\,b} \right)}^2} - \frac{1}{2}\,{{\left( {{{\tan }^{ - 1}}\,2} \right)}^2}} \right] \hfill \\ \hfill \\ evaluate\,\,the\,\,\lim it \hfill \\ \hfill \\ = \frac{1}{2}\,\left( {\frac{{{\pi ^2}}}{4} - \frac{{{\pi ^2}}}{{16}}} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{{3{\pi ^2}}}{{32}} \hfill \\ \hfill \\ \end{gathered}$