## Calculus: Early Transcendentals (2nd Edition)

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