Answer
$$\ln \left| x \right| - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^3} + x}}} \cr
& {\text{factor the denominator}} \cr
& = \int {\frac{{dx}}{{x\left( {{x^2} + 1} \right)}}} \cr
& {\text{The partial fraction decomposition of the integrand is}} \cr
& \frac{1}{{x\left( {{x^2} + 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} \cr
& {\text{Multiplying by }}x\left( {{x^2} + 1} \right){\text{ yields}} \cr
& 1 = A\left( {{x^2} + 1} \right) + x\left( {Bx + C} \right) \cr
& 1 = A{x^2} + A + B{x^2} + Cx \cr
& {\text{Collecting like powers of }}x{\text{, it becomes}} \cr
& 1 = A{x^2} + B{x^2} + Cx + A \cr
& {\text{Equating corresponding coefficients yields the following system }} \cr
& {\text{of linear equations}} \cr
& \,\,\,\,\,A + B = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A = 1 \cr
& {\text{Solving the system of linear equations we obtain}} \cr
& \,\,\,A = 1,\,\,\,B = - 1,\,\,\,C = 0 \cr
& \cr
& {\text{Then}}{\text{, the integrand can be written as}} \cr
& \frac{1}{{x\left( {{x^2} + 1} \right)}} = \frac{1}{x} - \frac{x}{{{x^2} + 1}} \cr
& \int {\frac{{dx}}{{x\left( {{x^2} + 1} \right)}}} = \int {\left( {\frac{1}{x} - \frac{x}{{{x^2} + 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \ln \left| x \right| - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + C \cr} $$
