GỢI Ý
$I=\int \dfrac{x^4 .x}{(x^2-1)\sqrt{x^2+1}}dx$ đặt $\sqrt{x^2+1}=t \Rightarrow x dx = tdt$
$I=\int \dfrac{(t^2-1)^2 .t}{(t^2 -2).t}dt=\int \dfrac{t^4-2t^2+1}{t^2-2}dt=\int \bigg (t^2 +\dfrac{1}{t^2-2} \bigg )dt$
Trong đó $I_1 =\dfrac{1}{t^2-2}dt= \dfrac{1}{2\sqrt 2}\int \bigg (\dfrac{1}{t-\sqrt 2 }-\dfrac{1}{t+\sqrt 2} \bigg )dt$
Dễ rồi nhé