Solve$\int\frac{x^4}{1-x^4}dx$ Question: Solve $\int\frac{x^4}{1-x^4}dx.$ 
My attempt: 
$$\int\frac{x^4}{1-x^4}dx = \int\frac{-(1-x^4)+1}{1-x^4}dx = \int 1 + \frac{1}{1-x^4}dx$$
To integrate $\int\frac{1}{1-x^4}dx,$ I apply substitution $x^2=\sin\theta.$ Then we have $2x \frac{dx}{d\theta} = \cos \theta.$ which implies that $\frac{dx}{d\theta}=\frac{\cos \theta}{2\sqrt{\sin \theta}}.$
So we have $\int \frac{1}{1-x^4}dx=\int\frac{1}{\cos^2\theta} \cdot \frac{\cos \theta}{2\sqrt{\sin \theta}} d\theta = \int\frac{1}{2\cos\theta \sqrt{\sin\theta}}d\theta.$ Then I stuck here. 
Any hint would be appreciated. 

By doing long division, you'll get $$-\int \left(-\frac{1}{2(x^2+1)}-\frac{1}{4(x+1)}+\frac{1}{4(x-1)}+1\right) dx$$
