Off by constants in integrating by parts $y-y\sqrt{y-2}$

Question

Please forgive me if these questions have already been asked, I tried my best to find similar ones in the archives.

I'm trying to integrate the above function, as follows: $$\int{y-y\sqrt{y-2}}\;\mathrm{d}y$$

WolframAlpha tells me the answer should be $$\frac{y^2}{2}-\frac{2(y-2)^\frac{5}{2}}{5}-\frac{4(y-2)^\frac{3}{2}}{3}$$ so I thought the integral of $-y\sqrt{y-2}$ should be $$\frac{2(y-2)^\frac{5}{2}}{5}+\frac{4(y-2)^\frac{3}{2}}{3}$$ (by the sum rule for integration), but it's not the answer I get when I ask WolframAlpha for that integral alone. It gives: $$\frac{2(y-2)^\frac{3}{2}(3y+4)}{15}$$

That really confuses me. Where does that come from?

I can get somewhat close to the last two terms in WolframAlpha's solution for the entire integral using integration by parts, but my constants are off. I have no idea how you would get $$\frac{2(y-2)^\frac{3}{2}(3y+4)}{15}$$

Here is my work for integrating $-y\sqrt{y-2}$ by parts: $$\int{-y\sqrt{y-2}}\;\mathrm{d}y$$ $$-\int{y\sqrt{y-2}}\;\mathrm{d}y$$ $$u=-y,\;\mathrm{d}v=\sqrt{y-2}$$ $$-\frac{2y\,(y-2)^\frac{3}{2}}{3}-\int{-\frac{2(y-2)^\frac{3}{2}}{3}}\mathrm\:{d}y$$ $$-\frac{2y\,(y-2)^\frac{3}{2}}{3}+\frac{4(y-2)^\frac{5}{2}}{15}$$

Thank you!

Answer 1

Your answer has 2 fractions with denominators 5 and 3, while WolframAlpha is giving an answer which is a single fraction with denominator 15. That suggests you should combine the 2 fractions in your answer, because that will definitely give you the 15 in the denominator, and then compare.

Answer 2

HINT:

$$-\frac25(y-2)^{5/2}-\frac43(y-2)^{3/2}=-(y-2)^{3/2}\left(\frac25(y-2)+\frac43\right)$$

SPOLIER ALERT Scroll over the highlighted area to reveal the solution

We have $-(y-2)^{3/2}\left(\frac25(y-2)+\frac43\right)=-\frac2{15}(y-2)^{3/2}\left(3y-4\right)$ as expected!

Answer 3

If you're willing to use Wolfram Alpha, just look at this search:
http://www.wolframalpha.com/input/?i=4%2F15+%28y-2%29%5E%285%2F2%29+-+%282y%29%2F3+%28y-2%29%5E%283%2F2%29

If you scroll down to alternate forms you will see that the two results are the same; therefore, you have a correct solution. (Ted is correct in that basic algebra in converting the fractions to a common denominator proves this result.)