Calculus: Early Transcendentals (2nd Edition)

$= \frac{1}{2}\,\,\left[ {\ln \frac{5}{2}} \right] + \frac{\pi }{4} - {\tan ^{ - 1}}(2)$
$\begin{gathered} \int_0^2 {\frac{x}{{{x^2} + 4x + 8}}dx} \hfill \\ \hfill \\ complete\,\,the\,\,square\,\,and\,\,factor \hfill \\ \hfill \\ \int_0^2 {\frac{x}{{{x^2} + 4x + 4 + 4}}dx} \hfill \\ \hfill \\ = \int_0^2 {\frac{x}{{\,{{\left( {x + 2} \right)}^2} + 4}}dx} \hfill \\ \hfill \\ set\,\,\,\left( {x + 2} \right) = u\, \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ x = u - 2\,\,\,\,\,\, \to \,\,\,\,\,\,dx = du \hfill \\ \hfill \\ {\text{changing}}\,\,{\text{the}}\,\,{\text{limits}} \hfill \\ \hfill \\ lower\,limit = \,0 + 2 = 2 \hfill \\ upper\,\,limit = 2 + 2 = 4 \hfill \\ \hfill \\ \,integral\,\,becomes \hfill \\ \hfill \\ \int_0^2 {\frac{x}{{\,{{\left( {x + 2} \right)}^2} + 4}}dx} = \int_2^4 {\frac{{u - 2}}{{{u^2} + 4}}du} \hfill \\ \hfill \\ split\,\,the\,\,numerator \hfill \\ \hfill \\ = \int_2^4 {\frac{u}{{{u^2} + 4}}du} \int_2^4 {\frac{u}{{{u^2} + 4}}du} \hfill \\ \hfill \\ \,Integrate\,and\,evaluate\, \hfill \\ \hfill \\ = \,\,\left[ {\frac{1}{2}\ln t} \right]_8^{20} = \frac{1}{2}\,\,\left[ {\ln \frac{5}{2}} \right] \hfill \\ \hfill \\ for - 2\int_2^4 {\frac{{du}}{{{u^2} + 4}}} \hfill \\ \hfill \\ {\text{Using}}\,\,\int_{}^{} {\frac{{dx}}{{{x^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\frac{x}{a} \hfill \\ \hfill \\ \,Integrate\,and\,evaluate\, \hfill \\ \hfill \\ = - 2\,\,\left[ {\frac{1}{2}{{\tan }^{ - 1}}\frac{u}{2}} \right]_2^4 = - \,\left( {{{\tan }^{ - 1}}2 - {{\tan }^{ - 1}}1} \right) = \frac{\pi }{4} - {\tan ^{ - 1}}2 \hfill \\ \hfill \\ {\text{Therefore}}{\text{,}} \hfill \\ \hfill \\ \int_0^2 {\frac{x}{{\,{{\left( {x + 2} \right)}^2} + 4}}dx} = \int_2^4 {\frac{{u - 2}}{{{u^2} + 4}}du} = \frac{1}{2}\,\,\left[ {\ln \frac{5}{2}} \right] + \frac{\pi }{4} - {\tan ^{ - 1}}(2) \hfill \\ \hfill \\ \end{gathered}$