Đặt \(\sqrt{x-1}-1=t\Rightarrow x=t^2+2t+2\Rightarrow dx=2\left(t+1\right)dt\) ; \(\left\{{}\begin{matrix}x=5\Rightarrow t=1\\x=10\Rightarrow t=2\end{matrix}\right.\)
\(F=\int\limits^2_1\dfrac{2\left(t+1\right)dt}{t^2+2t+2-2\left(t+1\right)}=2\int\limits^2_1\left(\dfrac{1}{t}+\dfrac{1}{t^2}\right)dt=2\left(lnt-\dfrac{1}{t}\right)|^2_1=2ln2+1\)