Answer
$$\frac{3}{2}\ln \left( {{x^2} + 1} \right) - \frac{1}{{2\left( {{x^2} + 1} \right)}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{3{x^3} + 4x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx \cr
& {\text{Decomposing the integrand into partial fractions}} \cr
& \frac{{3{x^3} + 4x}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{Ax + B}}{{{x^2} + 1}} + \frac{{Cx + D}}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{ }}\left( {\bf{1}} \right) \cr
& 3{x^3} + 4x = \left( {Ax + B} \right)\left( {{x^2} + 1} \right) + Cx + D \cr
& {\text{Expanding}} \cr
& 3{x^3} + 4x = A{x^3} + Ax + B{x^2} + B + Cx + D \cr
& 3{x^3} + 4x = A{x^3} + B{x^2} + Ax + Cx + B + D \cr
& {\text{Comparing coefficients}} \cr
& A = 3 \cr
& B = 0 \cr
& A + C = 4 \to C = 4 - A = 4 - 3 = 1 \cr
& B + D = 0 \to D = 0 \cr
& {\text{Substitute the constants into }}\left( {\bf{1}} \right) \cr
& \frac{{3{x^3} + 4x}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{3x + 0}}{{{x^2} + 1}} + \frac{{x + 0}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& \frac{{3{x^3} + 4x}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{3x}}{{{x^2} + 1}} + \frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& \int {\frac{{3{x^3} + 4x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx = \int {\frac{{3x}}{{{x^2} + 1}}} dx + \int {\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx \cr
& {\text{Rewrite integrands}} \cr
& = \frac{3}{2}\int {\frac{{2x}}{{{x^2} + 1}}} dx + \frac{1}{2}\int {\frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx \cr
& {\text{Integrate}} \cr
& {\text{ = }}\frac{3}{2}\ln \left| {{x^2} + 1} \right| + \frac{1}{2}\left( { - \frac{1}{{{x^2} + 1}}} \right) + C \cr
& {\text{ = }}\frac{3}{2}\ln \left( {{x^2} + 1} \right) - \frac{1}{{2\left( {{x^2} + 1} \right)}} + C \cr} $$
