Đặt \(\sqrt{x+1}=t\Rightarrow x=t^2-1\Rightarrow dx=2tdt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=1\\x=3\Rightarrow t=2\end{matrix}\right.\)
\(G=\int\limits^2_1\dfrac{t^2-1-3}{3t+t^2-1+3}.2tdt=2\int\limits^2_1\dfrac{t^2-2t}{t+1}dt=2\int\limits^2_1\left(t-3+\dfrac{3}{t+1}\right)dt\)
\(=\left(t^2-6t+6ln\left|t+1\right|\right)|^2_1=-3+6ln\left(\dfrac{3}{2}\right)\)