Answer
$$2{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{4xdx}}{{{x^3} + 4x}}} \cr
& {\text{Factor the denominator and simplify}} \cr
& \int {\frac{{4xdx}}{{{x^3} + 4x}}} = \int {\frac{{4xdx}}{{x\left( {{x^2} + 4} \right)}}} \cr
& = \int {\frac{{4dx}}{{{x^2} + 4}}} \cr
& = 4\int {\frac{{dx}}{{{x^2} + {{\left( 2 \right)}^2}}}} \cr
& {\text{Integrate using the formula }}\int {\frac{{dx}}{{{a^2} + {x^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + C \cr
& = 4\left( {\frac{1}{2}{{\tan }^{ - 1}}\left( {\frac{x}{2}} \right)} \right) + C \cr
& = 2{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr} $$
