Answer
\[ = \ln \left| {\frac{{x\,{{\left( {x - 2} \right)}^3}}}{{\,{{\left( {x + 2} \right)}^3}}}} \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{x^2} + 12x - 4}}{{{x^3} - 4x}}} dx \hfill \\
\hfill \\
{\text{factor}}\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\
\hfill \\
\frac{{{x^2} + 12x - 4}}{{{x^3} - 4x}} = \frac{{{x^2} + 12x - 4}}{{x\,\left( {{x^2} - 4} \right)}} = \frac{{{x^2} + 12x - 4}}{{x\,\left( {x + 2} \right)\left( {x - 2} \right)}} \hfill \\
\hfill \\
Using\,\,partial\,fractions \hfill \\
\hfill \\
= \frac{{{x^2} + 12x - 4}}{{x\,\left( {x + 2} \right)\,\left( {x - 2} \right)}} = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}} \hfill \\
\hfill \\
{\text{find}}\,\,{\text{the}}\,\,{\text{constants}} \hfill \\
\hfill \\
A = 1,\,\,B = - 3,\,\,C = 3 \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \frac{{{x^2} + 12x - 4}}{{x\,\left( {x + 2} \right)\,\left( {x - 2} \right)}} = - \,\frac{3}{{x + 2}}\,\, + \,\,\frac{1}{x}\,\, + \,\,\frac{3}{{x - 2}} \hfill \\
\hfill \\
\hfill \\
\int_{}^{} {\frac{{{x^2} + 12x - 4}}{{{x^3} - 4x}}} dx = \int_{}^{} {\,\left( { - \frac{3}{{x + 2}} + \frac{1}{x} + \frac{3}{{x - 2}}} \right)} dx \hfill \\
\hfill \\
{\text{integrate}} \hfill \\
\hfill \\
= - 3\ln \left| {x + 2} \right| + \ln \left| x \right| + 3\ln \left| {x - 2} \right| + C \hfill \\
\hfill \\
= \ln \left| {\,{{\left( {x + 2} \right)}^3}} \right| + \ln \left| x \right| + \ln \left| {\,{{\left( {x - 2} \right)}^3}} \right| + C \hfill \\
\hfill \\
= \ln \left| {\frac{{x\,{{\left( {x - 2} \right)}^3}}}{{\,{{\left( {x + 2} \right)}^3}}}} \right| + C \hfill \\
\end{gathered} \]
