## Calculus: Early Transcendentals (2nd Edition)

$= \frac{1}{2}\ln 2 - \frac{\pi }{4}$
$\begin{gathered} \int_{ - 1}^0 {\,\frac{x}{{{x^2} + 2x + 2}}\,\,dx} \,\,\, = \hfill \\ \hfill \\ completing\,\,the\,\,\,square \hfill \\ \hfill \\ \int_{ - 1}^0 {\frac{x}{{\,{{\left( {x + 1} \right)}^2} + 1}}} \,dx \hfill \\ \hfill \\ set\,\,x + 1\, = t\,\,then\,\,dx = dt \hfill \\ \hfill \\ = \int_0^1 {\frac{{\,\left( {t - 1} \right)}}{{{t^2} + 1}}\,dt\,} \hfill \\ \hfill \\ split\,\,the\,\,{\text{Integrand}} \hfill \\ \hfill \\ \int_0^1 {\,\,\left[ {\frac{t}{{{t^2} + 1}} - \frac{1}{{{t^2} + 1}}} \right]\,\,dt} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \int_0^1 {\frac{t}{{{t^2} + 1}}\,dt} - \int_0^1 {\frac{1}{{{t^2} + 1}}\,dt} \hfill \\ \hfill \\ {\text{Integrate}} \hfill \\ \hfill \\ \int_0^1 {\frac{t}{{{t^2} + 1}}} \,dt\,\, \hfill \\ \hfill \\ set\,\,{t^2} + 1 = u\,{\text{ }}then\,\,\,\,2tdt = du\,\,\,or\,\,tdt = \frac{{du}}{2} \hfill \\ \hfill \\ integral\,\,becomes \hfill \\ \hfill \\ \int_0^1 {\frac{t}{{{t^2} + 1}}} \,dt\,\, = \int_1^2 {\frac{{du}}{{2u}}} \hfill \\ \hfill \\ integrate\,and\,use\,\,the\,\,ftc \hfill \\ \hfill \\ = \frac{1}{2}\,\,\,\left[ {\ln u} \right]_1^2 = \frac{1}{2}\ln 2 \hfill \\ \hfill \\ for - \int_0^1 {\frac{1}{{{t^2} + 1}}dt} \hfill \\ \hfill \\ integrate\,and\,use\,\,the\,\,ftc \hfill \\ \hfill \\ = - \,\,\left[ {{{\tan }^{ - 1}}\,t} \right]_0^1 = - \,\left( {\frac{\pi }{4} - 0} \right) = - \frac{\pi }{4} \hfill \\ \hfill \\ \int_0^1 {\frac{t}{{{t^2} + 1}}\,dt} - \int_0^1 {\frac{1}{{{t^2} + 1}}\,dt} = \frac{1}{2}\ln 2 - \frac{\pi }{4} \hfill \\ \end{gathered}$