Answer
$2\tan^{-1} (\dfrac{x}{2})+c$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
Solve $\int \dfrac{4x}{x^3+4x}dx$
Re-arrange as: $\int \dfrac{4x}{x^3+4x}dx=\int \dfrac{dx}{(\dfrac{x}{2})^2+1}$
Plug in $\dfrac{x}{2}=a \implies dx=(2) da$
This implies that $\int \dfrac{dx}{(\dfrac{x}{2})^2+1}=\int \dfrac{2da}{a^2+1}=2 \tan^{-1} a+c=2\tan^{-1} (\dfrac{x}{2})+c$
