Đặt $t=\sqrt{x^2-1} $ thì $t^2=x^2-1$, suy ra:
$2tdt=2xdx ; \frac{xdx}{2x^2-1+3 \sqrt{x^2-1} }=\frac{xdx}{2(x^2-1)+3 \sqrt{x^2-1}-1 }=\frac{tdt}{2t^2+3t+1} $
Khi đó:
$\int\limits f(x)dx=\int\limits \frac{tdt}{2t^2+3t+1} $
Ta có:
$\frac{t}{2t^2+3t+1} =\frac{t}{(2t+1)(t+1)}=\frac{a}{2t+1}+\frac{b}{t+1} =\frac{(a+2b)t+a+b}{(2t+1)(t+1)} $
Đồng nhất đẳng thức, ta được:
$\begin{cases}a+2b=1 \\ a+b=0 \end{cases}\leftrightarrow \begin{cases}a=-1 \\ b=1 \end{cases} $
Khi đó:
$\frac{t}{2t^2+3t+1}=-\frac{1}{2t+1}+\frac{1}{t+1}$
Do đó:
$\int\limits f(x)dx=\int\limits (-\frac{1}{2t+1}+\frac{1}{t+1})dt=-\frac{1}{2}\ln|2t+1|+\ln|t+1|+C $
$=\frac{1}{2}\ln \frac{(t+1)^2}{|2t+1|} +C =\frac{1}{2}\ln \frac{(\sqrt{x^2-1}+1)^2}{2\sqrt{x^2-1}+1} +C $