# Very complicated algebraic simplification problem

So I solved an indefinite integral for Khan Academy multiple-choice problem. I got the right answer, but none of the multiple-choice answers seemed to fit.

On the left is what I got, on the right is the correct answer on the multiple-choice problem. The two expressions are actually equal, but I just don't understand the algebraic simplification necessary to get from the left side to the right side. Any help would be appreciated. $$\frac{2}{5} (x+2)^{5/2}-\frac{4}{3} (x+2)^{3/2}=\frac{1}{3} (2 x) (x+2)^{3/2}-\frac{4}{15} (x+2)^{5/2}$$

Note: WolframAlpha also confirms the equation is true.

• Since you want to verify that your integral and the book's is the same, you could also subtract one from the other. If your answer is a constant, then you know your answer is right. – Bernard Massé Apr 23 '18 at 18:53

The proper way to factor an expression is to take out the lowest power of common factors. Thus $$\frac 25(x+2)^{5/2} - \frac 43 (x+2)^{3/2} = (x+2)^{3/2} \left( \frac 25 (x+2) - \frac 43 \right)$$ and $$\frac 13 (2x) (x+2)^{3/2} - \frac 4{15}(x+2)^{5/2} = (x+2)^{3/2} \left( \frac 13 (2x) - \frac 4{15} (x+2) \right).$$
Now you just need to show $$\frac 25 (x+2) - \frac 43 = \frac 13 (2x) - \frac 4{15} (x+2).$$ Both sides easily reduce to $\dfrac 25 x - \dfrac 8{15}.$
• In other words, write $y=(x+2)^{\frac 1 2}$ (and rewrite the $2x$ as $2(x+2-2)$) and you have a regular old polynomial equation. – Jack M Apr 23 '18 at 18:04