Answer
$2\ln \left| x \right| + \frac{1}{2}\ln \left( {{x^2} + 4} \right) - \frac{1}{2}\arctan \left( {\frac{x}{2}} \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{3{x^2} - x + 8}}{{{x^3} + 4x}}} dx \cr
& {\text{Factoring the denominator}} \cr
& = \int {\frac{{3{x^2} - x + 8}}{{x\left( {{x^2} + 4} \right)}}} dx \cr
& {\text{The partial fraction decomposition of the integrand has the}} \cr
& {\text{form}} \cr
& \frac{{3{x^2} - x + 8}}{{x\left( {{x^2} + 4} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 4}} \cr
& {\text{Multiply both sides by the least common denominator}} \cr
& 3{x^2} - x + 8 = A\left( {{x^2} + 4} \right) + \left( {Bx + C} \right)x \cr
& 3{x^2} - x + 8 = A{x^2} + 4A + B{x^2} + Cx \cr
& {\text{Collecting like terms}} \cr
& 3{x^2} - x + 8 = \left( {A{x^2} + B{x^2}} \right) + Cx + 4A \cr
& {\text{We obtain the following system of equations}} \cr
& A + B = 3 \cr
& C = - 1 \cr
& 4A = 8 \cr
& A = 2,{\text{ }}B = 1,{\text{ }}C = - 1 \cr
& {\text{Therefore}}{\text{,}} \cr
& \frac{{3{x^2} - x + 8}}{{x\left( {{x^2} + 4} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 4}} \cr
& \frac{{3{x^2} - x + 8}}{{x\left( {{x^2} + 4} \right)}} = \frac{2}{x} + \frac{{x - 1}}{{{x^2} + 4}} \cr
& \int {\frac{{3{x^2} - x + 8}}{{x\left( {{x^2} + 4} \right)}}} dx = \int {\left( {\frac{2}{x} + \frac{{x - 1}}{{{x^2} + 4}}} \right)} dx \cr
& = \int {\left( {\frac{2}{x} + \frac{x}{{{x^2} + 4}} - \frac{1}{{{x^2} + 4}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = 2\ln \left| x \right| + \frac{1}{2}\ln \left( {{x^2} + 4} \right) - \frac{1}{2}\arctan \left( {\frac{x}{2}} \right) + C \cr} $$
