question in attachment
Hello, DINOCALC09!
Did you do the substitution?
(I agree with the Captain.)
If the substitution $\displaystyle u = \sqrt{x-1}$ is made, the integral .$\displaystyle \int^5_2\frac{\sqrt{x-1}}{x}\,dx\;=\$
We have: .$\displaystyle u \:=\: \sqrt{x-1}\quad\Rightarrow\quad x \:=\:u^2-1\quad\Rightarrow\quad dx \:=\;2u\,du$
Substitute: .$\displaystyle \int\frac{u}{u^2+1}(2u\,du) \;=\;\int\frac{2u^2}{u^2+1}\,du$
Change limits: .$\displaystyle \begin{Bmatrix}x = 2 & \Rightarrow & u = 1 \\ x = 5 & \Rightarrow & u = 2\end{Bmatrix}$
. . Answer: .$\displaystyle \int^2_1\frac{2u^2}{u+1}\,du$ . (E)