Answer
\[ = - \ln \left| {{x^4}} \right| + \ln \left| {\,{{\left( {x - 1} \right)}^5}} \right| + \frac{3}{{x - 1}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{x^2} - 4}}{{{x^3} - 2{x^2} + x}}} dx \hfill \\
\hfill \\
{\text{factor}}\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\
\hfill \\
\frac{{{x^2} - 4}}{{{x^3} - 2{x^2} + x}} = \frac{{{x^2} - 4}}{{x\left( {{x^2} - 2x + 1} \right)}} = \frac{{{x^2} - 4}}{{x{{\left( {x - 1} \right)}^2}}} \hfill \\
\hfill \\
Applying\,\,partial\,fractions \hfill \\
\hfill \\
\frac{{{x^2} - 4}}{{x{{\left( {x - 1} \right)}^2}}} = \frac{A}{x} + \frac{B}{{x - 1}} + \frac{C}{{{{\left( {x - 1} \right)}^2}}} \hfill \\
\hfill \\
find\,\,the\,\,consta\,nts \hfill \\
\hfill \\
A = - 4,{\text{ B}} = 5,{\text{ C}} = - 3 \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_{}^{} {\frac{{{x^2} - 4}}{{{x^3} - 2{x^2} + x}}} dx = \int_{}^{} {\,\left( { - \frac{4}{x} + \frac{5}{{x - 1}} - \frac{3}{{\,{{\left( {x - 1} \right)}^2}}}} \right)} dx \hfill \\
\hfill \\
integrating \hfill \\
\hfill \\
= - 4\ln + 5\ln \left| {x - 1} \right| + \frac{3}{{x - 1}} + C \hfill \\
\hfill \\
= - \ln \left| {{x^4}} \right| + \ln \left| {\,{{\left( {x - 1} \right)}^5}} \right| + \frac{3}{{x - 1}} + C \hfill \\
\hfill \\
\end{gathered} \]
