Đặt: $t=\sqrt{x-1}\Rightarrow t^2=x-1 \Rightarrow 2tdt=dx$
Ta có:
$\int\limits_1^2\dfrac{x}{1+\sqrt{x-1}}dx$
$=\int\limits_0^1\dfrac{(t^2+1)2tdt}{1+t}$
$=\int\limits_0^1\dfrac{2t^3+2t}{t+1}dt$
$=\int\limits_0^1\left(2t^2-2t+4-\dfrac{4}{t+1}\right)dt$
$=\left(\dfrac{2t^3}{3}-t^2+4t-4\ln(t+1)\right)\left|\begin{array}{l}1\\0\end{array}\right.$
$=\dfrac{11}{3}-4\ln2$