Integral of $\left(\frac{x^4}{1-x^4}\right)^k$ I'm stumped. How does one solve the following expression:
$$\int\left(\frac{x^4}{1-x^4}\right)^kdx, k\in\mathbb{N}$$
I thought it would be a fun challenge, but it beats me.

You can obtain a recurrence relation and then find a formula using sums and products for a closed form solution. $k\geq 1$, let 
$$ I_k(x) = \int \left( \frac{x^4}{1-x^4} \right) ^k {\rm d}x$$
Using by parts with $u = (1-x^4)^{-k}$ and ${\rm d}v = x^{4k} {\rm d}x$, we will get
$$
\begin {align}
I_k(x) &=  \frac{x^{4k+1}}{(4k+1)(1-x^4)^k} - \frac{4k}{4k+1} \int \frac{x^{4(k+1)}}{(1-x^4)^{k+1}} {
\rm d}x
\\ & = \frac{x^{4k+1}}{(4k+1)(1-x^4)^k} - \frac{4k}{4k+1} I_{k+1}(x)
\end {align}
$$
Rewriting, we get
$$
I_{k+1}(x) =  \frac{x^{4k+1}}{4k(1-x^4)^k} - \frac{4k+1}{4k} I_k(x)
$$
Now, getting a formula for $I_k$ in terms of sums might be a little tedious but shouldn't be too difficult. First solve it for $k=1$ using partial fractions,
$$ I_1(x) = \frac12 \arctan(x) +\frac14 \ln \left| \frac{1+x}{1-x} 
\right| -x + C$$
Then, for $k \geq 2$,

$$
I_k(x) = I_1(x) \prod_{n=1}^{k-1} \left(-\frac{4n+1}{4n} \right) + \sum_{n=1}^{k-1} \left( \prod_{m=n+1}^{k-1} -\frac{4m+1}{4m}  \right)\frac{x^{4n+1}}{4n(1-x^4)^n} +C
$$

