Đặt \(\sqrt[]{x+1}=t\Rightarrow x=t^2-1\Rightarrow dx=2t.dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=1\\x=3\Rightarrow t=2\end{matrix}\right.\)
\(I=2\int\limits^2_1\dfrac{\left(t^2-1\right)^2.2t.dt}{t^2.t}=4\int\limits^2_1\dfrac{\left(t^4-2t^2+1\right)}{t^2}dt=4\int\limits^2_1\left(t^2-2+\dfrac{1}{t^2}\right)dt\)
\(=4\left(\dfrac{1}{3}t^3-2t-\dfrac{1}{t}\right)|^2_1=\dfrac{10}{3}\)