## Calculus: Early Transcendentals (2nd Edition)

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