Calculus: Early Transcendentals (2nd Edition)

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