## Thomas' Calculus 13th Edition

$$- 2\ln \left| {v - 1} \right| + \ln \left| {v - 2} \right| + \ln \left| {v - 3} \right| + C$$
\eqalign{ & \int {\frac{{\left( {3v - 7} \right)dv}}{{\left( {v - 1} \right)\left( {v - 2} \right)\left( {v - 3} \right)}}} \cr & {\text{Decompose the integrand into partial fractions.}} \cr & {\text{The form of the partial fraction decomposition is}} \cr & \frac{{3v - 7}}{{\left( {v - 1} \right)\left( {v - 2} \right)\left( {v - 3} \right)}} = \frac{A}{{v - 1}} + \frac{B}{{v - 2}} + \frac{C}{{v - 3}} \cr & \cr & {\text{Multiplying by }}\left( {v - 1} \right)\left( {v - 2} \right)\left( {v - 3} \right){\text{, we have}} \cr & 3v - 7 = A\left( {v - 2} \right)\left( {v - 3} \right) + B\left( {v - 1} \right)\left( {v - 3} \right) + C\left( {v - 1} \right)\left( {v - 2} \right)\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{if we set }}v = 1{\text{ into equation }}\left( {\bf{1}} \right) \cr & 3\left( 1 \right) - 7 = A\left( {1 - 2} \right)\left( {1 - 3} \right) + B\left( 0 \right) + C\left( 0 \right) \cr & A = \frac{{ - 4}}{2} = - 2 \cr & \cr & {\text{if we set }}v = 2{\text{ into equation }}\left( {\bf{1}} \right) \cr & 3\left( 2 \right) - 7 = A\left( 0 \right) + B\left( {2 - 1} \right)\left( {2 - 3} \right) + C\left( 0 \right) \cr & B = \frac{{ - 1}}{{ - 1}} = 1 \cr & \cr & {\text{if we set }}v = 3{\text{ into the equation }}\left( {\bf{1}} \right) \cr & 3\left( 3 \right) - 7 = A\left( 0 \right) + B\left( 0 \right) + C\left( {3 - 1} \right)\left( {3 - 2} \right) \cr & C = \frac{2}{2} = 1 \cr & \cr & {\text{The decomposition of the integrand is}} \cr & \frac{{3v - 7}}{{\left( {v - 1} \right)\left( {v - 2} \right)\left( {v - 3} \right)}} = - \frac{2}{{v - 1}} + \frac{1}{{v - 2}} + \frac{1}{{v - 3}} \cr & \cr & \int {\frac{{\left( {3v - 7} \right)dv}}{{\left( {v - 1} \right)\left( {v - 2} \right)\left( {v - 3} \right)}}} = - \int {\frac{2}{{v - 1}}} dv + \int {\frac{1}{{v - 2}}} dv + \int {\frac{1}{{v - 3}}dv} \cr & {\text{Integrating, we get}} \cr & = - 2\ln \left| {v - 1} \right| + \ln \left| {v - 2} \right| + \ln \left| {v - 3} \right| + C \cr}