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I'm trying to calculate the following integral :$\int \sqrt x (2-3x^2)^2dx$, but somehow I can't get it right, here's what I did:

First I expanded the expression:

\begin{align}&\int \sqrt x (4-12x^2 + 9x^4) \\&= \int(4\sqrt x-12x^2\sqrt x + 9x^4\sqrt x) \end{align}

Then I evaluated each term individually:

\begin{align}&4 \int \sqrt x - 12 \int x ^\frac{5}{2} + 9 \int x^\frac{9}{2} \\& = \frac{4x^\frac{3}{2} \cdot 2}{3} - \frac{12x^\frac{7}{2} \cdot 2}{7} + \frac{9x^\frac{11}{2} \cdot 2}{11}\end{align}

Which gives

$$\frac{8x^\frac{3}{2}}{3} - \frac{24x^\frac{7}{2}}{7} + \frac{18x^\frac{11}{2}}{11}$$

What's wrong with my solution?

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    $\begingroup$ What gave the idea that there's something wrong with your solution? $\endgroup$ Commented Dec 17, 2017 at 21:46
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    $\begingroup$ There is no mistake in your calculation except for plus $C$ integration constant :) $\endgroup$ Commented Dec 17, 2017 at 21:49
  • $\begingroup$ You forgot a minus sign in $4 * \int \sqrt x + 12 * \int x ^\frac{5}{2} + 9 * \int x^\frac{9}{2}$, but remembered it in the next line (it's just a little typo). $\endgroup$ Commented Dec 17, 2017 at 22:01
  • $\begingroup$ @JoséCarlosSantos this wolframalpha.com/input/?i=integrate+sqrt(x)(2-3x%5E2)%5E2 $\endgroup$ Commented Dec 17, 2017 at 22:22

1 Answer 1

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You just forget the constant (and the minus sign as commented), other than that, your answer is correct.

We can verify the solution by differentiating the solution.

\begin{align}\frac{d}{dx}\left(\frac{8x^\frac{3}{2}}{3} - \frac{24x^\frac{7}{2}}{7} + \frac{18x^\frac{11}{2}}{11}+c\right) &=\left( \frac32\right)\frac{8x^\frac12}{3}-\left( \frac72\right)\frac{24x^\frac52}{7}+\left(\frac{11}2\right) \frac{18x^\frac92}{11}\\&= 4x^\frac12-12x^\frac52+9x^\frac92\\&= \sqrt x (4-12x^2 + 9x^4) \end{align}

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