Integral $\int \frac{\tan^4 \theta \,d \theta}{1-\tan^2 \theta}$? I have to evaluate this indefinite integral $$\int \frac{\tan^4 \theta\, d \theta}{1-\tan^2 \theta}$$
I tried it as follows $$I=\int\frac{(\sec^2 \theta-1)\tan^2 \theta\, d \theta}{1-\tan^2 \theta}=\int\frac{\sec^2 \theta \tan^2 \theta\, d \theta}{1-\tan^2 \theta}-\int\frac{\tan^2 \theta \,d \theta}{1-\tan^2 \theta}$$
First part of integration can be easily solved by substitution but how to solve the second part? Help to solve it by other method if you have. Thanks!

Here is another way to proceed:
$\displaystyle\int\frac{\tan^4\theta}{1-\tan^2\theta}d\theta=\int\frac{\tan^4\theta-1}{1-\tan^2\theta}d\theta+\int\frac{1}{1-\tan^2\theta}d\theta=-\int(\tan^2\theta+1)d\theta+\int\frac{\cos^2\theta}{\cos^2\theta-\sin^2\theta}d\theta$
$=\displaystyle-\int\sec^2\theta \;d\theta+\int\frac{\frac{1}{2}(1+\cos 2\theta)}{\cos 2\theta}d\theta= -\tan\theta+\int\left(\frac{1}{2}\sec2\theta+\frac{1}{2}\right)d\theta$
$=\displaystyle-\tan\theta+\frac{1}{4}\ln\big|\sec2\theta+\tan2\theta\big|+\frac{1}{2}\theta+C$
