<s>[INST] 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 $
$u = x^2 , du = 2x dx, \frac{du}{2}=xdx$
$\frac{1}{2}\int \frac{1}{u^2+4}du$
 [/INST] Use $$\frac{x}{x^4+4}=\frac{x}{x^4+4x^2+4-4x^2}=\frac{x}{(x^2-2x+2)(x^2+2x+2)}=$$
$$=\frac{1}{4}\left(\frac{1}{x^2-2x+2}-\frac{1}{x^2+2x+2}\right)=\frac{1}{4}\left(\frac{1}{(x-1)^2+1}-\frac{1}{(x+1)^2+1}\right).$$
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