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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.

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  • $\begingroup$ 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. $\endgroup$ – Bernard Massé Apr 23 '18 at 18:53
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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 $\dfrac 25 x - \dfrac 8{15}.$

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  • $\begingroup$ 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. $\endgroup$ – Jack M Apr 23 '18 at 18:04
  • $\begingroup$ Excellent! What I was missing was the part about taking out the lowest power of common factors. I'm refreshing my calculus skills because it's been about 12 years and I've lost much of it, but I'm getting it all back; sometimes the algebra is still a bit rough. $\endgroup$ – Alex Pilafian Apr 23 '18 at 18:10

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