Let's take an integral: $\int \frac{4\tan^3x-1}{2\tan x+1}dx$
Let $u=\tan(\frac{x}{2})$, then:
$\tan x = \frac{2t}{1-t^2}$
$dx=\frac{2}{1+t^2}$
I get $\int \frac{\frac{64t^3}{(1-t^2)^3}-1}{\frac{4t}{1-t^2}-1}dt$. And, well, this doesn't look hospitable. Is there any better solution?
On
Hint:
By long division
$$\dfrac{4\tan^3x-1}{2\tan x+1}=2\tan^2x-\tan x+\dfrac{\tan x+1}{2\tan x+1}$$
Now $$\dfrac{\tan x+1}{2\tan x+1}=\dfrac{\sin x+\cos x}{2\sin x+\cos x}$$
Write $$\sin x+\cos x=a(2\sin x+\cos x)+b\cdot\dfrac{d(2\sin x+\cos x)}{dx}$$
Compare the coefficients of $\sin x,\cos x$ to find $a,b$
Let $\tan{x}=t$.
Thus, we need to get $$\int\frac{4t^3-1}{(2t+1)(1+t^2)}dt.$$