Đặt: $\sqrt{x+1}=t \Rightarrow x+1=t^2 \Rightarrow dx=2tdt$
Đổi cận: $x=0 \Rightarrow t=1$
$x=3 \Rightarrow t=2$
Ta có:
$\int\limits_0^3\dfrac{x^2+2}{\sqrt{x+1}}dx$
$=\int\limits_1^2\dfrac{(t^2-1)^2+2}{t}2tdt$
$=2\int\limits_1^2(t^4-2t^2+3)dt$
$=\left(\dfrac{2t^5}{5}-\dfrac{4t^3}{3}+6t\right)\left|\begin{array}{l}2\\1\end{array}\right.=\dfrac{136}{15}$