TITLE: Good substitution for this integral
QUESTION [5 upvotes]: What is $$\int \frac{4t}{1-t^4}dt$$ is there some kind of substitution which might help .Note that here $t=\tan(\theta)$

REPLY [3 votes]: Doesn't this work?
$$\int \frac{4t}{1-t^4}dt=\int \frac{2t}{1+t^2}-\frac{-2t}{1-t^2} dt=\ln |\frac{1+t^2}{1-t^2}|+C$$
Also $t=\tan \theta$ substitution works quite well. $$\int \frac{4\tan \theta}{1-\tan^4 \theta}\sec^2 \theta d \theta=\int \frac {4 \tan \theta }{1-\tan^2 \theta} d\theta=2\int \frac{1+\tan \theta}{1-\tan \theta}-\frac{1-\tan \theta}{1+\tan \theta} d\theta$$
$$2\int \frac{1+\tan \theta}{1-\tan \theta}-\frac{1-\tan \theta}{1+\tan \theta} d\theta=2\int\frac{cos \theta+\sin \theta}{\cos \theta-\sin \theta}+\frac{\cos \theta-\sin \theta}{cos \theta+\sin \theta}d\theta$$
Note that $(\sin \theta+\cos \theta)'=\cos \theta-\sin \theta$, $(\cos \theta-\sin \theta)'=-\sin \theta-\cos \theta$.
I think you can continue from here.