Answer
$$ - \frac{{{x^2}}}{2} - 2\ln \left| {4 - {x^2}} \right| + \frac{1}{2}\ln \left| {2 + x} \right| - \frac{1}{2}\ln \left| {2 - x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3} + 2}}{{4 - {x^2}}}} dx \cr
& \cr
& {\text{Use long division }}\frac{{{x^3} + 2}}{{4 - {x^2}}} = - x + \frac{{4x + 2}}{{4 - {x^2}}} \cr
& \frac{{{x^3} + 2}}{{4 - {x^2}}} = - x + \frac{{4x}}{{4 - {x^2}}} + \frac{2}{{4 - {x^2}}} \cr
& \cr
& {\text{Decomposing }}\frac{2}{{4 - {x^2}}}{\text{ into partial fractions}} \cr
& \frac{2}{{4 - {x^2}}} = \frac{2}{{\left( {2 + x} \right)\left( {2 - x} \right)}} = \frac{A}{{2 + x}} + \frac{B}{{2 - x}} \cr
& 2 = A\left( {2 - x} \right) + B\left( {2 + x} \right) \cr
& {\text{If }}x = - 2 \to A = \frac{1}{2} \cr
& {\text{If }}x = 2 \to B = \frac{1}{2} \cr
& {\text{Then}}{\text{, }}\frac{2}{{4 - {x^2}}} = \frac{{1/2}}{{2 + x}} + \frac{{1/2}}{{2 - x}} \cr
& \cr
& - x + \frac{{4x}}{{4 - {x^2}}} + \frac{2}{{4 - {x^2}}} = - x + \frac{{4x}}{{4 - {x^2}}} + \frac{1}{{2\left( {2 + x} \right)}} + \frac{1}{{2\left( {2 - x} \right)}} \cr
& \cr
& \int {\frac{{{x^3} + 2}}{{4 - {x^2}}}} dx = \int {\left( { - x + \frac{{4x}}{{4 - {x^2}}} + \frac{1}{{2\left( {2 + x} \right)}} + \frac{1}{{2\left( {2 - x} \right)}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = - \frac{{{x^2}}}{2} - 2\ln \left| {4 - {x^2}} \right| + \frac{1}{2}\ln \left| {2 + x} \right| - \frac{1}{2}\ln \left| {2 - x} \right| + C \cr} $$
