I want to integrate

$$\int x^2 \sqrt{x^3 +1}~dx$$

I tried it with integration by parts (because we have a product here), but an online calculator did it with integration by substition.

Would this still be correct?

$$\frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int \frac{1}{3} x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} \cdot 3~dx \\ = \frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} ~dx\\ = \frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \frac{1}{4} x^4\cdot 2(x^3 +1)^{-\frac{1}{2}} \\ = \frac{1}{3}x^3 \sqrt{x^3+1} - \frac{1}{2} x^4 \frac{1}{\sqrt{x^3+1}}$$

I think this is wrong because when $x=1$ I get a different result than when I insert $x=1$ into

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Can someone tell me where I went wrong and why we rather use integration by substition instead of integration by parts here?

  • $\begingroup$ Problem is here: $\frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int \frac{1}{3} x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} \cdot 3$. That should be $3x^2$. $\endgroup$ Jul 27, 2022 at 13:55
  • $\begingroup$ @eyeballfrog Ahh I missed that. Thanks! $\endgroup$ Jul 27, 2022 at 13:59
  • $\begingroup$ Do you know how to integrate it otherwise, without integrating by parts? $\endgroup$ Jul 27, 2022 at 13:59
  • $\begingroup$ @insipidintegrator No, because I didn't know one should substitute $x^3 +1$ when using integration by substitution here. $\endgroup$ Jul 27, 2022 at 14:02

3 Answers 3


$$\frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int \frac{1}{3} x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} \cdot 3 \tag{1}$$

In your first line, you miss $x^2$, which is from the chain rule. It should be

$$\frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int \frac{1}{3} x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} \cdot 3x^2$$

Here is a trick, note that $dx^3=d(x^3+1)$, so we have

$$\begin{align} \int x^2 \sqrt{x^3 +1}~dx&=\frac{1}{3}\int \sqrt{x^3 +1}~d(x^3+1)\tag{2}\\ \\ &=\frac{1}{3}\cdot\frac{2}3\cdot (x^3+1)^{3/2}+C \end{align}$$

  • $\begingroup$ Can you elaborate on that quick way please? How do you get to $\frac{1}{3}$? $\endgroup$ Jul 27, 2022 at 14:03
  • $\begingroup$ $d(x^3+1)=3x^2dx$ so you need factor $1/3$ to cancel it. $\endgroup$
    – MathFail
    Jul 27, 2022 at 14:05
  • $\begingroup$ You mean because $dx = \frac{1}{3} \frac{du}{x^2}$? What happens with the $x^2$ then? $\endgroup$ Jul 27, 2022 at 14:06
  • $\begingroup$ In my Eq.(2), you do derivative on RHS and compare with LHS, then you will know why $1/3$ there $\endgroup$
    – MathFail
    Jul 27, 2022 at 14:08

The first line of your calculation is missing a factor of $x^2$, which arises when using the chain rule to compute $d(\sqrt{x^3 + 1})$.

Instead, applying integration by parts with $$u = \sqrt{x^3 + 1}, \qquad dv = x^2$$ gives $$\left(\frac{1}{3} x^3\right) \left(\sqrt{x^3 + 1}\right) - \int \left(\frac{1}{3} x^3\right) \left( \frac{3 x^2 \,dx}{2 \sqrt{x^3 + 1}} \right) = \frac{\sqrt{x^3 + 1}}{3 x^3} - \frac{1}{2} \int \frac{x^5 \,dx}{\sqrt{x^3 + 1}} .$$ At this point the remaining integral is arguably worse than the one we started with, but we can still evaluate it using the same substitution the online calculator used, $$w = x^3 + 1, \qquad dw = 3 x^2;$$ we have $$\int \frac{x^5 \,dx}{\sqrt{x^3 + 1}} = \int \frac{[(x^3 + 1) - 1] \cdot x^2 \,dx}{\sqrt{x^3 + 1}} = \frac{1}{3} \int \frac{(w - 1) \,dw}{\sqrt{w}},$$ and the lattermost integrand is a sum of power functions. But it's easier just to apply that substitution to the original integral instead: $$\int x^2 \sqrt{x^3 + 1} \,dx = \frac{1}{3} \int \sqrt{w} \,dw .$$


$\int x^2 \sqrt{x^3 +1}~dx = \frac{2}{9}(x^3+1)^{\frac{3}{2}} +c$ by direct integration


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