a) \(\dfrac{{t + 3}}{{{t^3} + {t^2} - 2t}} = \dfrac{{t + 3}}{{t\left( {{t^2} + t - 2} \right)}}\) \( = \dfrac{{t + 2}}{{t\left( {t - 1} \right)\left( {t + 2} \right)}} + \dfrac{1}{{t\left( {t - 1} \right)\left( {t + 2} \right)}}\)
\( = \dfrac{1}{{t\left( {t - 1} \right)}} + \dfrac{1}{3}\left( {\dfrac{1}{{t\left( {t - 1} \right)}} - \dfrac{1}{{t\left( {t + 2} \right)}}} \right)\) \( = \dfrac{1}{{t - 1}} - \dfrac{1}{t} + \dfrac{1}{3}\left[ {\dfrac{1}{{t - 1}} - \dfrac{1}{t} - \dfrac{1}{2}\left( {\dfrac{1}{t} - \dfrac{1}{{t + 2}}} \right)} \right]\)
\( = \dfrac{1}{{t - 1}} - \dfrac{1}{t} + \dfrac{1}{{3\left( {t - 1} \right)}} - \dfrac{1}{{3t}} - \dfrac{1}{{6t}} - \dfrac{1}{{3\left( {t + 2} \right)}}\) \( = \dfrac{4}{{3\left( {t - 1} \right)}} - \dfrac{3}{{2t}} - \dfrac{1}{{3\left( {t + 2} \right)}}\) \( \Rightarrow \int {\dfrac{{t + 3}}{{{t^3} + {t^2} - 2t}}dt} = \int {\left( {\dfrac{4}{{3\left( {t - 1} \right)}} - \dfrac{3}{{2t}} - \dfrac{1}{{3\left( {t + 2} \right)}}} \right)dt} \)
\( = \dfrac{4}{3}\ln \left| {t - 1} \right| - \dfrac{3}{2}\ln \left| t \right| - \dfrac{1}{3}\ln \left| {t + 2} \right| + C\)
