Answer
$$\frac{1}{6}\ln \left( {\frac{{59}}{{36}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{\left( {v + 1} \right)\left( {v + 2} \right)}}{{2{v^3} + 9{v^2} + 12v + 36}}} dv \cr
& {\text{multiply in the numerator}} \cr
& \int_0^1 {\frac{{{v^2} + 3v + 2}}{{2{v^3} + 9{v^2} + 12v + 36}}} dv \cr
& {\text{set }}u = 2{v^3} + 9{v^2} + 12v + 36{\text{ then }}du = \left( {6{v^2} + 18v + 12} \right)dv \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,du = 6\left( {{v^2} + 3v + 2} \right)dv \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{6}du = \left( {{v^2} + 3v + 2} \right)dv \cr
& {\text{switch the limits of integration}} \cr
& u = 2{v^3} + 9{v^2} + 12v + 36,\,\,\,\,v = 1 \to u = 59 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,v = 0 \to u = 36 \cr
& {\text{use the change of variable}} \cr
& \int_0^1 {\frac{{{v^2} + 3v + 2}}{{2{v^3} + 9{v^2} + 12v + 36}}} dv = \int_{36}^{59} {\frac{{\left( {1/6} \right)du}}{u}} \cr
& = \frac{1}{6}\int_{36}^{59} {\frac{{du}}{u}} \cr
& {\text{integrate}} \cr
& = \frac{1}{6}\left[ {\ln \left| u \right|} \right]_{36}^{59} \cr
& {\text{evaluate the limits}} \cr
& = \frac{1}{6}\left[ {\ln \left| {59} \right| - \ln \left| {36} \right|} \right] \cr
& {\text{logarithmic properties}} \cr
& = \frac{1}{6}\ln \left( {\frac{{59}}{{36}}} \right) \cr} $$
