Answer
$$ - \frac{3}{8}\ln \left| v \right| + \frac{1}{{16}}\ln \left| {v + 2} \right| + \frac{5}{{16}}\ln \left| {v - 2} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{v + 3}}{{2{v^3} - 8v}}} dv \cr
& {\text{Decompose the integrand into partial fractions.}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{{v + 3}}{{2{v^3} - 8v}} = \frac{{v + 3}}{{2v\left( {{v^2} - 4} \right)}} = \frac{{v + 3}}{{2v\left( {v + 2} \right)\left( {v - 2} \right)}} \cr
& \frac{{v + 3}}{{2v\left( {v + 2} \right)\left( {v - 2} \right)}} = \frac{A}{{2v}} + \frac{B}{{v + 2}} + \frac{C}{{v - 2}} \cr
& \cr
& {\text{Multiplying by }}2v\left( {v + 2} \right)\left( {v - 2} \right){\text{, we have}} \cr
& v + 3 = A\left( {v + 2} \right)\left( {v - 2} \right) + 2Bv\left( {v - 2} \right) + 2Cv\left( {v + 2} \right)\,\,\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{if we set }}v = 0{\text{ into equation }}\left( {\bf{1}} \right) \cr
& 0 + 3 = A\left( 2 \right)\left( { - 2} \right) + B\left( 0 \right) + C\left( 0 \right) \cr
& A = - \frac{3}{4} \cr
& \cr
& {\text{if we set }}v = - 2{\text{ into equation }}\left( {\bf{1}} \right) \cr
& - 2 + 3 = A\left( 0 \right) + 2B\left( { - 2} \right)\left( { - 4} \right) + C\left( 0 \right) \cr
& B = \frac{1}{{16}} \cr
& \cr
& {\text{if we set }}v = 2{\text{ into equation }}\left( {\bf{1}} \right) \cr
& 2 + 3 = A\left( 0 \right) + 2B\left( 0 \right) + 2C\left( 2 \right)\left( {2 + 2} \right) \cr
& C = \frac{5}{{16}} \cr
& \cr
& {\text{The decomposition of the integrand is}} \cr
& \frac{{v + 3}}{{2v\left( {v + 2} \right)\left( {v - 2} \right)}} = \frac{{ - 3/4}}{{2v}} + \frac{{1/16}}{{v + 2}} + \frac{{5/16}}{{v - 2}} \cr
& \cr
& \int {\frac{{v + 3}}{{2{v^3} - 8v}}} dv = - \frac{3}{8}\int {\frac{1}{v}dv + \frac{1}{{16}}\int {\frac{1}{{v + 2}}dv} + \frac{5}{{16}}\int {\frac{1}{{v - 2}}} dv} \cr
& {\text{Integrating, we get}} \cr
& = - \frac{3}{8}\ln \left| v \right| + \frac{1}{{16}}\ln \left| {v + 2} \right| + \frac{5}{{16}}\ln \left| {v - 2} \right| + C \cr} $$
