Solve$\int\frac{x^4}{1-x^4}dx$

Question

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.

Answer 1

HINT:$$\frac{1}{1-x^4}=\frac{1}{2}\left(\frac{1}{1+x^2}+\frac{1}{1-x^2}\right)$$

You handle these. I guess?

Answer 2

Hint. Alternatively, one may use a partial fraction decomposition to get $$ \frac{1}{x^4-1}=\frac{1}{4(x-1)}-\frac{1}{4(x+1)}-\frac{1}{2 \left(x^2+1\right)} $$ then one may integrate each term easily.

Answer 3

Hint: $$\frac{1}{1-x^4}=\frac{1}{2}\left(\frac{1-x^2+1+x^2}{1-x^4}\right)=\frac{1}{2(1+x^2)}+\frac{1}{2(1-x^2)}$$

Answer 4

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