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