Answer
$$ - \frac{3}{8}\ln \left| x \right| + \frac{1}{{16}}\ln \left| {x + 2} \right| + \frac{5}{{16}}\ln \left| {x - 2} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{x + 3}}{{2{x^3} - 8x}}} dx \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{x + 3}}{{2{x^3} - 8x}} = \frac{{x + 3}}{{2x\left( {{x^2} - 4} \right)}} = \frac{{x + 3}}{{2x\left( {x + 2} \right)\left( {x - 2} \right)}} \cr
& \frac{{x + 3}}{{2x\left( {x + 2} \right)\left( {x - 2} \right)}} = \frac{A}{{2x}} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}} \cr
& {\text{Multiply by }}2x\left( {x + 2} \right)\left( {x - 2} \right) \cr
& x + 3 = A\left( {x + 2} \right)\left( {x - 2} \right) + 2Bx\left( {x - 2} \right) + 2Cx\left( {x + 2} \right)\,\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{If we set }}x = 0,{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& \,\,\,\,\,\,\,\,3 = A\left( 2 \right)\left( { - 2} \right) + 2B\left( 0 \right) + 2C\left( 0 \right) \cr
& \,\,\,\,\,\,\,\,A = - \frac{3}{4} \cr
& \cr
& {\text{If we set }}x = - 2,{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& \,\,\,\,\,\,\,\,1 = A\left( 0 \right) + 2B\left( { - 2} \right)\left( { - 4} \right) + 2C\left( 0 \right) \cr
& \,\,\,\,\,\,\,\,B = \frac{1}{{16}} \cr
& \cr
& {\text{If we set }}x = 2,{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& \,\,\,\,\,\,\,\,5 = A\left( 0 \right) + 2B\left( 0 \right) + 2C\left( 2 \right)\left( 4 \right) \cr
& \,\,\,\,\,\,\,\,C = \frac{5}{{16}} \cr
& \cr
& {\text{Substituting }}A,B,C \cr
& \frac{{x + 3}}{{2x\left( {x + 2} \right)\left( {x - 2} \right)}} = \frac{A}{{2x}} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}} = \frac{{ - 3/4}}{{2x}} + \frac{{1/16}}{{x + 2}} + \frac{{5/16}}{{x - 2}} \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \int {\frac{{x + 3}}{{2{x^3} - 8x}}} dx = \int {\left( {\frac{{ - 3/4}}{{2x}} + \frac{{1/16}}{{x + 2}} + \frac{{5/16}}{{x - 2}}} \right)} dx \cr
& = - \frac{3}{8}\ln \left| x \right| + \frac{1}{{16}}\ln \left| {x + 2} \right| + \frac{5}{{16}}\ln \left| {x - 2} \right| + C \cr} $$
