okay, so the problem is:
integral from 2 to 6 y/(y-2)^1/2 dy
i know you have to start by saying lim a-->2+ integral from a to 6 y/(y-2)^1/2
but then i'm kind of blanking on how to integrate/go from there
$\displaystyle \int \frac{y}{\sqrt{y-2}} \, dy$
let $\displaystyle u = y - 2$
$\displaystyle du = dy$
$\displaystyle y = u + 2$
$\displaystyle \int \frac{u+2}{\sqrt{u}} \, du$
$\displaystyle \int u^{\frac{1}{2}} + 2u^{-\frac{1}{2}} \, du$
integrate, back substitute, then evaluate your improper integral