<s>[INST] Surface Area by Integration $$2\pi\int_{3}^6\left(\frac{1}{3}x^\frac{3}{2}-x^\frac{1}{2}\right)\left(1+\left(\frac{1}{2}x^\frac{1}{2}-\frac{1}{2}x^\frac{-1}{2}\right)^2\right)^\frac{1}{2}dx$$
I've managed to simplify this down to the equation below (not sure if it'll help), but I still can't integrate it. 
Please help.
$$\frac{\pi}{3}\int_{3}^6(x^4-4x^3-2x^2+12x+9)^\frac{1}{2}dx$$
PS. The answer you should get is 9$\pi$.
 [/INST] You did a good bit of the hard work yourself, but these other computations may help:
\begin{align}
\frac{\pi}{3}\int_3^6\sqrt{x^4-4x^3-2x^2+12x+9}\;dx &= \frac{\pi}{3}\int_3^6 (x^2-2x-3)\;dx\tag{simplify}\\[1em]
                  &= \frac{\pi}{3}\int_3^6 x^2\;dx - \frac{2\pi}{3}\int_3^6x\;dx -\pi\int_3^61\;dx\\[1em]
                  &= \left.\frac{\pi x^3}{9}\right|_3^6-\frac{2\pi}{3}\int_3^6x\;dx-\pi\int_3^61\;dx\tag{FTC}\\[1em]
                  &= 21\pi-\frac{2\pi}{3}\int_3^6x\;dx-\pi\int_3^61\;dx\\[1em]
                  &= 21\pi+\left.\left(-\frac{\pi x^2}{3}\right)\right|_3^6-\pi\int_3^61\;dx\tag{FTC}\\[1em]
                  &= 12\pi-\pi\int_3^61\;dx\\[1em]
                  &= 12\pi+\left.(-\pi x)\right|_3^6\\[1em]
                  &= 9\pi.
\end{align}
Hence, you can see the integral evaluates to $9\pi$, as desired.
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