Answer
\[ = 4\ln \left| x \right| + 2\ln \left( {{x^2} + 8} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{8\,\left( {{x^2} + 4} \right)}}{{x\,\left( {{x^2} + 8} \right)}}dx} \hfill \\
\hfill \\
Applying\,\,partial\,\,fractions \hfill \\
\hfill \\
\frac{{8\,\left( {{x^2} + 4} \right)}}{{x\,\left( {{x^2} + 8} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 8}} \hfill \\
\hfill \\
{\text{find}}\,\,{\text{the}}\,\,{\text{constants}} \hfill \\
\hfill \\
A = 1,{\text{ }}B = 4,{\text{ C}} = 0 \hfill \\
\hfill \\
\frac{{8\,\left( {{x^2} + 4} \right)}}{{x\,\left( {{x^2} + 8} \right)}} = \frac{4}{x} + \frac{{4x}}{{{x^2} + 8}} \hfill \\
\hfill \\
Therefore, \hfill \\
\hfill \\
\int_{}^{} {\frac{{8\,\left( {{x^2} + 4} \right)}}{{x\,\left( {{x^2} + 8} \right)}}dx} = \int_{}^{} {\,\left( {\frac{4}{x} + \frac{{4x}}{{{x^2} + 8}}} \right)} dx \hfill \\
\hfill \\
integrating \hfill \\
\hfill \\
= 4\ln \left| x \right| + \int_{}^{} {\frac{{4x}}{{{x^2} + 8}}} \hfill \\
\hfill \\
let\,\,u = {x^2} + 8 \hfill \\
\hfill \\
= 4\ln \left| x \right| + \int_{}^{} {\frac{{2u}}{u}} \hfill \\
\hfill \\
= 4\ln \left| x \right| + 2\ln u + C \hfill \\
\hfill \\
substitute\,\,back\,\,u = {x^2} + 8 \hfill \\
\hfill \\
= 4\ln \left| x \right| + 2\ln \left( {{x^2} + 8} \right) + C \hfill \\
\end{gathered} \]
