Đặt \(\sqrt{x+2}=t\Rightarrow x=t^2-2\Rightarrow dx=2tdt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=\sqrt{2}\\x=1\Rightarrow t=\sqrt{3}\end{matrix}\right.\)
\(B=\int\limits^{\sqrt{3}}_{\sqrt{2}}\left(t^2-2\right).t.2t.dt=\int\limits^{\sqrt{3}}_{\sqrt{2}}\left(2t^4-4t^2\right)dt=\left(\dfrac{2}{5}t^5-\dfrac{4}{3}t^3\right)|^{\sqrt{3}}_{\sqrt{2}}=...\)



