## Calculus: Early Transcendentals (2nd Edition)

$= \frac{1}{6}\ln \frac{1}{6}$
$\begin{gathered} \int_0^5 {\frac{2}{{{x^2} - 4x + 32}}dx} \hfill \\ \hfill \\ Applying\,\,partial\,fractions \hfill \\ \hfill \\ \frac{2}{{{x^2} - 4x + 32}} = \frac{2}{{\,\left( {x - 8} \right)\,\left( {x + 4} \right)}} = \frac{A}{{x - 8}} + \frac{B}{{x + 4}} \hfill \\ \hfill \\ find\,\,A{\text{ and B}} \hfill \\ \hfill \\ A = \frac{1}{6}\,\,{\text{ and }}\,B = - \frac{1}{6} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{2}{{{x^2} - 4x + 32}} = - \frac{1}{{6\,\left( {x + 4} \right)}} + \frac{1}{{6\,\left( {x - 8} \right)}} \hfill \\ \hfill \\ integrating \hfill \\ \hfill \\ \int_0^5 {\frac{2}{{{x^2} - 4x + 32}}dx} = \int_0^5 {\,\left( { - \frac{1}{{6\,\left( {x + 4} \right)}} + \frac{1}{{6\,\left( {x - 8} \right)}}} \right)} dx \hfill \\ \hfill \\ = \,\left. {\left( { - \frac{1}{6}\ln \left| {x + 4} \right| + \frac{1}{6}\ln \left| {x - 8} \right|} \right)} \right|_0^5 \hfill \\ \hfill \\ evaluate\,\,\,the\,\,{\text{limits}} \hfill \\ \hfill \\ = \,\left( { - \frac{1}{6}\ln \left| {5 + 4} \right| + \frac{1}{6}\ln \left| {5 - 8} \right|} \right) - \,\left( { - \frac{1}{6}\ln \left| {0 + 4} \right| + \frac{1}{6}\ln \left| {0 - 8} \right|} \right) \hfill \\ \hfill \\ = \,\left( { - \frac{1}{6}\ln \left| 9 \right| + \frac{1}{6}\ln \left| 3 \right|} \right) - \,\left( { - \frac{1}{6}\ln \left| 4 \right| + \frac{1}{6}\ln \left| 8 \right|} \right) \hfill \\ \hfill \\ = \frac{1}{6}\ln \frac{1}{9} - \frac{1}{6}\ln \left| {\frac{8}{4}} \right| \hfill \\ \hfill \\ solution \hfill \\ \hfill \\ = \frac{1}{6}\ln \frac{1}{6} \hfill \\ \end{gathered}$