$$I=\int_0^1 \dfrac{x^2dx}{(x+1)\sqrt{x+1}}=\int_0^1 \dfrac{x^2-1}{(x+1)\sqrt{x+1}}dx+\int_0^1 \dfrac{dx}{\sqrt{(x+1)^3}} \\
=I_1+I_2 \\
I_1=\int_0^1 \dfrac{(x-1)(x+1)}{(x+1)\sqrt{x+1}}=\int_0^1 \dfrac{x-1}{\sqrt{x+1}}dx \\ $$
$\text{đặt }t=\sqrt{x+1} \rightarrow t^2=x+1 \rightarrow 2tdt=dx \\
\text{đổi cận } x=0 \rightarrow t=1 , x=1 \rightarrow t=\sqrt{2} \\ $
$$I_1=\int_1^{\sqrt{2}} \dfrac{2(t^2-2)tdt}{t}=\int_0^{\sqrt{2}} 2(t^2-2)dt \\
I_2=\int_0^1 \dfrac{dx}{\sqrt{(x+1)^3}} \\ $$
$\text{đặt }u=\sqrt{x+1} \rightarrow u^2=x+1 \rightarrow 2udu=dx \\
\text{đổi cận }x=0 \rightarrow t=1 , x=1 \rightarrow t=\sqrt{2} $
$$I_2=\int_1^{\sqrt{2}} \dfrac{2udu}{u^3}=\int_1^{\sqrt{2}} \dfrac{2du}{u^2} \\$$
$\text{đến đây quá dễ để mà tính}$