Given that $\displaystyle y=(3x+2)\sqrt{x+2}$ and $\displaystyle \frac{dy}{dx}=\frac{3x+8}{2\sqrt{x+2}}$.
Integrate $\displaystyle \int^7_{2}\frac{x}{2\sqrt{x+2}}$ with respect to x
First off, as HallsofIvy point out, the substitution u = x+2 or x = u - 2 will give you two integrals that you can easily integrate (and the way I personally would have done this problem). In what follows, the constants of integration will be supressed. First, using the fact that
$\displaystyle \displaystyle \int \dfrac{9x+14}{2\sqrt{x+2}}dx = (3x+2)\sqrt{x+2}$
then
$\displaystyle \displaystyle 9 \int \dfrac{x}{\sqrt{x+2}}dx + 14 \int \dfrac{1}{\sqrt{x+2}}dx = (6x+4)\sqrt{x+2},\;\;(1)$
noting that what you want is the first integral. Also $\displaystyle 9x+14 = 9(x+2) - 4$ so your integral can be written as
$\displaystyle \displaystyle \int \dfrac{9(x+2)-4}{\sqrt{x+2}}dx = 9 \int \sqrt{x+2}\,dx - 4\int \dfrac{1}{\sqrt{x+2}}dx$
so
$\displaystyle \displaystyle 9 \int \sqrt{x+2}\,dx - 4\int \dfrac{1}{\sqrt{x+2}}dx = (6x+4)\sqrt{x+2},\;\;\;(2)$
Now use integration by parts $\displaystyle \int u dv = uv - \int v du$ on the first integral in (2)
$\displaystyle u = \sqrt{x+2}, \;\; du = \dfrac{dx}{2\sqrt{x+2}}$
$\displaystyle dv = dx,\;\;\; v = x$
gives
$\displaystyle \displaystyle 9 x \sqrt{x+2} - \dfrac{9}{2}\int \dfrac{x}{\sqrt{x+2}}dx - 4\int \dfrac{1}{\sqrt{x+2}}dx = (6x+4)\sqrt{x+2}$
or
$\displaystyle \displaystyle - \dfrac{9}{2}\int \dfrac{x}{\sqrt{x+2}}dx - 4\int \dfrac{1}{\sqrt{x+2}}dx = (-3x+4)\sqrt{x+2}$
or
$\displaystyle \displaystyle 9\int \dfrac{x}{\sqrt{x+2}}dx + 8\int \dfrac{1}{\sqrt{x+2}}dx = (6x-8)\sqrt{x+2}.\;\;(3)$
Now multiply $\displaystyle 7 \times (3) - 4 \times (1)$ gives
$\displaystyle \displaystyle 27 \int \dfrac{x}{\sqrt{x+2}}dx = 18(x-4)\sqrt{x+2}$
from which it follows that
$\displaystyle \displaystyle \int \dfrac{x}{\sqrt{x+2}}dx = \frac{2}{3} (x-4)\sqrt{x+2}$.
Kinda long and again, not the way I would have done this problem.