## Calculus: Early Transcendentals (2nd Edition)

$= 2\ln \frac{{3\,\left( 2 \right)}}{4} - \frac{4}{9}$
$\begin{gathered} \int_1^3 {\frac{2}{{{t^3}\,\left( {t + 1} \right)}}} dt \hfill \\ \hfill \\ Applying\,\,partial\,fraction \hfill \\ \hfill \\ \frac{2}{{{t^3}\,\left( {t + 1} \right)}} = \frac{{{A_1}}}{t} + \frac{{{A_2}}}{{{t^2}}} + \frac{{{A_3}}}{{{t^3}}} + \frac{{{A_4}}}{{t + 1}} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \frac{2}{t} - \frac{2}{{{t^2}}} + \frac{2}{{{t^3}}} - \frac{2}{{t + 1}} \hfill \\ \hfill \\ integrating \hfill \\ \hfill \\ \int_1^3 {\frac{2}{{{t^3}\,\left( {t + 1} \right)}}dt} = \int_1^3 {\,\left( {\frac{2}{t} - \frac{2}{{{t^2}}} + \frac{2}{{{t^3}}} - \frac{2}{{t + 1}}} \right)dt} \hfill \\ \hfill \\ \left. { = \,\left( {2\ln \left| t \right| + \frac{2}{t} - \frac{1}{{{t^2}}} - 2\ln \left| {t + 1} \right|} \right)} \right|_1^3 \hfill \\ \hfill \\ {\text{evaluate}}\,\,{\text{the}}\,\,{\text{limits}} \hfill \\ \hfill \\ = \,\left( {2\ln \left| 3 \right| + \frac{2}{3} - \frac{1}{{{3^2}}} - 2\ln \left| {3 + 1} \right|} \right) - \,\left( {2\ln \left| 1 \right| + \frac{2}{1} - \frac{1}{{{1^2}}} - 2\ln \left| {1 + 1} \right|} \right) \hfill \\ \hfill \\ = \,\left( {2\ln 3 + \frac{2}{3} - \frac{1}{9} - 2\ln 4 - 2 + 1 + 2\ln 2} \right) \hfill \\ \hfill \\ solution \hfill \\ \hfill \\ = 2\ln \frac{{3\,\left( 2 \right)}}{4} - \frac{4}{9} \hfill \\ \end{gathered}$