Answer
\[ = \ln \frac{{{{\left| {x - 2} \right|}^2}}}{{\left| {x - 3} \right|}} - \frac{6}{{\,\left( {x - 3} \right)}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{x^2} - x}}{{\,\left( {x - 2} \right)\,{{\left( {x - 3} \right)}^2}}}dx} \hfill \\
\hfill \\
Applying\,\,partial\,fraction \hfill \\
\hfill \\
\frac{{{x^2} - x}}{{\,\left( {x - 2} \right)\,{{\left( {x - 3} \right)}^2}}} = \frac{A}{{\,\left( {x - 2} \right)}} + \frac{B}{{\,\left( {x - 3} \right)}} + \frac{C}{{\,{{\left( {x - 3} \right)}^2}}} \hfill \\
\hfill \\
find\,\,the\,\,constan ts \hfill \\
\hfill \\
A = 2,{\text{ B}} = - 1,{\text{ C}} = 6 \hfill \\
\hfill \\
= \frac{2}{{\,\left( {x - 2} \right)}} - \frac{1}{{\,\left( {x - 3} \right)}} + \frac{6}{{\,{{\left( {x - 3} \right)}^2}}} \hfill \\
\hfill \\
integrating \hfill \\
\hfill \\
\int_{}^{} {\frac{{{x^2} - x}}{{\,\left( {x - 2} \right)\,{{\left( {x - 3} \right)}^2}}}dx} = \int_{}^{} {\,\left( {\frac{2}{{\,\left( {x - 2} \right)}} - \frac{1}{{\,\left( {x - 3} \right)}} + \frac{6}{{\,{{\left( {x - 3} \right)}^2}}}} \right)dx} \hfill \\
\hfill \\
= 2\ln \left| {x - 2} \right| - \ln \left| {x - 3} \right| + 6\,\,\left[ {\frac{{\,{{\left( {x - 3} \right)}^{ - 2 + 1}}}}{{ - 2 + 1}}} \right] + C \hfill \\
\hfill \\
= \ln {\left| {x - 2} \right|^2} - \ln \left| {x - 3} \right| - \frac{6}{{\,\left( {x - 3} \right)}} + C \hfill \\
\hfill \\
= \ln \frac{{{{\left| {x - 2} \right|}^2}}}{{\left| {x - 3} \right|}} - \frac{6}{{\,\left( {x - 3} \right)}} + C \hfill \\
\hfill \\
\hfill \\
\hfill \\
\end{gathered} \]
