# Math Help - integral

1. ## integral

question in attachment

2. Originally Posted by ThePerfectHacker
None of them.
E?

RonL

3. Hello, DINOCALC09!

Did you do the substitution?
(I agree with the Captain.)

If the substitution $u = \sqrt{x-1}$ is made, the integral . $\int^5_2\frac{\sqrt{x-1}}{x}\,dx\;=\$

We have: . $u \:=\: \sqrt{x-1}\quad\Rightarrow\quad x \:=\:u^2-1\quad\Rightarrow\quad dx \:=\;2u\,du$

Substitute: . $\int\frac{u}{u^2+1}(2u\,du) \;=\;\int\frac{2u^2}{u^2+1}\,du$

Change limits: . $\begin{Bmatrix}x = 2 & \Rightarrow & u = 1 \\ x = 5 & \Rightarrow & u = 2\end{Bmatrix}$

. . Answer: . $\int^2_1\frac{2u^2}{u+1}\,du$ . (E)