Integrate $\int \frac{1}{x^4+4}dx$ 
Integrate $\displaystyle \int \dfrac{1}{x^4+4}dx$. 

I could try breaking this up into two quadratic trinomials, but that seems like it would be a lot of work. If that is the best way here how do I do it most efficiently?
 A: I did it like this. You can Complete it Easily after this.

A: Notice
$$\frac{1}{x^4+4} = \frac{1}{(x^2+2)^2 - 4x^2}
= \frac{1}{(x^2+2x+2)(x^2-2x+2)}$$
Since 
$$\begin{align}
\frac{1}{x^2-2x+2} - \frac{1}{x^2+2x+2} &= \frac{4x}{(x^2+2x+2)(x^2-2x+2)}\\
\frac{1}{x^2-2x+2} + \frac{1}{x^2+2x+2} &= \frac{2x^2+4}{(x^2+2x+2)(x^2-2x+2)}
\end{align}$$
We have
$$\frac{1}{(x^2+2x+2)(x^2-2x+2)}
= \frac14\left[\frac{1-\frac{x}{2}}{x^2-2x+2} + \frac{1+\frac{x}{2}}{x^2+2x+2}\right]\\
= \frac18\left[\frac{x+2}{x^2+2x+2} - \frac{x-2}{x^2-2x+2}\right]
= \frac18\left[\frac{(x+1)+1}{(x+1)^2+1} - \frac{(x-1)-1}{(x-1)^2+1}\right]
$$
Up to an integration constant, this give us
$$\begin{align}
\int\frac{dx}{x^4+1}
&= \frac{1}{16}\log\left(\frac{(x+1)^2+1}{(x-1)^2+1}\right)
+ \frac18 \left[\tan^{-1}(x+1)+\tan^{-1}(x-1)\right]\\
&= \frac{1}{16}\log\left(\frac{(x+1)^2+1}{(x-1)^2+1}\right)
+ \frac18 \left[\tan^{-1}(x+1)+\tan^{-1}(x-1)\right]\\
&=\frac{1}{16}\log\left(\frac{(x+1)^2+1}{(x-1)^2+1}\right)
+ \frac18 \tan^{-1}\left(\frac{2x}{2-x^2}\right)
\end{align}
$$
A: If you really want to avoid partial fractions, you can do it this way:
Expand in Taylor series (convergent for $|x| < \sqrt{2}$):
$$ \dfrac{1}{4+x^4} = \dfrac{1/4}{1 + x^4/4} = \sum_{n=0}^\infty \dfrac{(-1)^n}{4^{n+1}} x^{4n} $$
Integrate term-by-term
$$ \int \dfrac{1}{4+x^4} = \sum_{n=0}^\infty \dfrac{(-1)^n}{4^{n+1}} \dfrac{x^{4n+1}}{4n+1} = \dfrac{x}{16} \text{LerchPhi}\left(-\dfrac{x^4}{4},1,\dfrac{1}{4}\right)$$
where 
$$ \text{LerchPhi}(z,a,v) = \sum_{n=0}^\infty \dfrac{z^n}{(n+v)^a}$$
Of course you might want to convert that LerchPhi expression to something
more elementary...
