Answer
$\ln |\dfrac{(v-2)(v-3)}{(v-1)^2}|+c$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
Solve $\int \dfrac{3v-7}{(v-1)(v-2)(v-3)}dv$
Write the integral into partial fractions as follows: $\int \dfrac{3v-7}{(v-1)(v-2)(v-3)}dv=\int [\dfrac{-2}{v-1}+\dfrac{1}{v-2}+\dfrac{1}{v-3}]dv$
This implies that $-2\ln |v-1|+\ln |v-2|+\ln |v-3|+c=\ln |\dfrac{(v-2)(v-3)}{(v-1)^2}|+c$
