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