## Thomas' Calculus 13th Edition

$$- \frac{1}{2}\ln \left| t \right| + \frac{1}{6}\ln \left| {t + 2} \right| + \frac{1}{3}\ln \left| {t - 1} \right| + C$$
\eqalign{ & \int {\frac{{dt}}{{{t^3} + {t^2} - 2t}}} \cr & {\text{Decompose the integrand into partial fractions}} \cr & \frac{1}{{{t^3} + {t^2} - 2t}} = \frac{1}{{t\left( {{t^2} + t - 2} \right)}} = \frac{1}{{t\left( {t + 2} \right)\left( {t - 1} \right)}} \cr & \frac{1}{{t\left( {t + 2} \right)\left( {t - 1} \right)}} = \frac{A}{t} + \frac{B}{{t + 2}} + \frac{C}{{t - 1}} \cr & {\text{Multiply by }}t\left( {t + 2} \right)\left( {t - 1} \right) \cr & \cr & 1 = A\left( {t + 2} \right)\left( {t - 1} \right) + Bt\left( {t - 1} \right) + Ct\left( {t + 2} \right)\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{If we set }}t = 0,{\text{ into the equation }}\left( {\bf{1}} \right) \cr & \,\,\,\,\,\,\,\,1 = A\left( 2 \right)\left( { - 1} \right) + B\left( 0 \right) + C\left( 0 \right) \cr & \,\,\,\,\,\,\,\,A = - \frac{1}{2} \cr & \cr & {\text{If we set }}t = - 2,{\text{ into the equation }}\left( {\bf{1}} \right) \cr & \,\,\,\,\,\,\,\,1 = A\left( 0 \right) + B\left( { - 2} \right)\left( { - 3} \right) + C\left( 0 \right) \cr & \,\,\,\,\,\,\,\,B = \frac{1}{6} \cr & \cr & {\text{If we set }}t = 1,{\text{ into the equation }}\left( {\bf{1}} \right) \cr & \,\,\,\,\,\,\,\,1 = A\left( 0 \right) + B\left( 0 \right) + C\left( 1 \right)\left( {1 + 2} \right) \cr & \,\,\,\,\,\,\,\,C = \frac{1}{3} \cr & \cr & {\text{Substituting }}A,B,C \cr & \frac{1}{{t\left( {t + 2} \right)\left( {t - 1} \right)}} = \frac{A}{t} + \frac{B}{{t + 2}} + \frac{C}{{t - 1}} = \frac{{ - 1/2}}{t} + \frac{{1/6}}{{t + 2}} + \frac{{1/3}}{{t - 1}} \cr & \cr & {\text{Therefore}}{\text{,}} \cr & \int {\frac{{dt}}{{{t^3} + {t^2} - 2t}}} = \int {\left( {\frac{{ - 1/2}}{t} + \frac{{1/6}}{{t + 2}} + \frac{{1/3}}{{t - 1}}} \right)} dt \cr & = - \frac{1}{2}\ln \left| t \right| + \frac{1}{6}\ln \left| {t + 2} \right| + \frac{1}{3}\ln \left| {t - 1} \right| + C \cr}