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 $
$u = x^2 , du = 2x dx, \frac{du}{2}=xdx$
$\frac{1}{2}\int \frac{1}{u^2+4}du$
 A: Use $$\frac{x}{x^4+4}=\frac{x}{x^4+4x^2+4-4x^2}=\frac{x}{(x^2-2x+2)(x^2+2x+2)}=$$
$$=\frac{1}{4}\left(\frac{1}{x^2-2x+2}-\frac{1}{x^2+2x+2}\right)=\frac{1}{4}\left(\frac{1}{(x-1)^2+1}-\frac{1}{(x+1)^2+1}\right).$$
A: Starting from
$$\int \frac{x}{x^4+4}\,dx$$
Make the substitution $u=x^2$. Then, $\dfrac{du}{2}=x\,dx$ and the integral becomes
$$\frac{1}{2}\int \frac{1}{u^2+4}\,du$$
Now let $u=2\tan(\theta)$. Then, $du=2\sec^2(\theta)\,d\theta$ and the integral becomes
$$\frac{1}{4}\int \frac{\sec^2(\theta)}{1+\tan^2{(\theta)}}\,d\theta$$
and since $1+\tan^2{(\theta)}=\sec^2(\theta)$, we may rewrite the integral as
$$\frac{1}{4}\int d\theta=\frac{1}{4}\theta+C$$
where $\theta=\arctan\left(\dfrac{u}{2}\right)$ and $u=x^2$, so we conclude that
$$\frac{1}{4}\theta+C=\frac{1}{4}\arctan\left(\frac{u}{2}\right)+C=\frac{1}{4}\arctan\left(\frac{x^2}{2}\right)+C$$
A: $$I=\int \frac{x}{x^4+4}dx= \frac{1}{2} \int \frac{dt}{4+t^2}=\frac{1}{2} \tan^{-1} t/2+C=\frac{1}{4} \tan^{-1} x^2/2+C$$ Here, we took $x^2=t \implies 2xdx=dt$
