Integrate $\int \frac{x}{x^4+4}dx$ I'm having trouble solving the integral
$\int \frac{x}{x^4+4}dx$
I already tried the following but, I'm getting stuck after step 3.
$\int \frac{x}{(x^2)^2+4}dx $
$u = x^2 , du = 2x dx, \frac{du}{2}=xdx$
$\frac{1}{2}\int \frac{1}{u^2+4}du$

$$I=\int \frac{x}{x^4+4}dx= \frac{1}{2} \int \frac{dt}{4+t^2}=\frac{1}{2} \tan^{-1} t/2+C=\frac{1}{4} \tan^{-1} x^2/2+C$$ Here, we took $x^2=t \implies 2xdx=dt$
