$$\frac{2a^4x}{x^4+4a^4}=\frac{A}{x-a(1+i)}+\frac{B}{x-a(1-i)}+\frac{C}{x+a(1+i)}+\frac{D}{x+a(1-i)}$$
I am trying to calculate this, but it is too complicated. I am thinking that maybe I am using wrong method. What is the easiest method to solve this?
On
Just for completeness, I'll do this using the method I mentioned in the comment above (even though this is an old question). Use $u=x^2\implies du=2xdx$ to get $$\int\dfrac{2a^4x}{x^4+4a^4}dx=\int\dfrac{a^4}{u^2+4a^4}du$$ Now, use the substitution $u=2a^2\tan(t)\implies du=2a^2\sec^2(t)dt$ to get $$\int\dfrac{a^4}{u^2+4a^4}du=\int\dfrac{a^4}{(2a^2\tan(t))^2+4a^4}2a^2\sec^2(t)dt=\int\dfrac{a^4}{4a^4(\tan^2(t)+1)}2a^2\sec^2(t)dt=\frac{1}{2}a^2\int 1dt=\frac{1}{2}a^2t+C=\frac{1}{2}a^2\arctan\Bigl(\frac{u}{2a^2}\Bigr)+C=\boxed{\frac{1}{2}a^2\arctan\Bigl(\frac{x^2}{2a^2}\Bigr)+C}$$
You may easily get rid of the $a$ parameter and just compute the partial fraction decomposition of $f(z)=\frac{z}{z^4+4}$. Such meromorphic function has simple poles at $\pm 1\pm i$, hence $$ \frac{z}{z^4+4} = \frac{A_+^+}{z-(+1+i)}+\frac{A_+^-}{z-(+1-i)}+\frac{A_-^+}{z-(-1+i)}+\frac{A_-^-}{z-(-1-i)} $$ with $$ A_+^+ = \text{Res}\left(f(z),z=+1+i\right)=\lim_{z\to (1+i)}\frac{z(z-(1+i))}{z^4+4}\stackrel{dH}{=}\lim_{z\to(1+i)}\frac{2z-(1+i)}{4z^3} $$ that is a straightforward limit to compute, and the same argument applies to $A_{\pm}^\pm$.