Đặt \(1+\sqrt{x-1}=t\Rightarrow x=t^2-2t+2\Rightarrow dx=\left(2t-2\right)dt\) ; \(\left\{{}\begin{matrix}x=1\Rightarrow t=1\\x=2\Rightarrow t=2\end{matrix}\right.\)
\(C=\int\limits^2_1\dfrac{\left(t^2-2t+2\right).\left(2t-2\right)dt}{t}=2\int\limits^2_1\left(t^2-3t+4-\dfrac{2}{t}\right)dt\)
\(=2\left(\dfrac{t^3}{3}-\dfrac{3t^2}{2}+4t-2ln\left|t\right|\right)|^2_1=\dfrac{11}{3}-4ln2\)