I am trying to solve this integral $$\int\frac{x\cdot \sqrt[3]{x+2}}{x+\sqrt[3]{x+2}} dx$$ I can do it by brute force (means using a substitution then long division and then substitutions again) but it's too long (suspiciously long solution). Is there any better way to solve it? I guess it should be. If you could provide me with the process that leads to the answer that would really help.

  • 2
    $\begingroup$ From the looks of this, I'm not sure this integral can be solved "easily." $\endgroup$
    – Jared
    Commented May 19, 2013 at 1:33
  • 1
    $\begingroup$ I don't know whether this is any better than what you already have (since you haven't told us what exactly you already have), but I'd substitute $u=\root3\of{x+2}$, $x=u^3-2$, $dx=3u^2\,du$ to turn it into a rational function, and take it from there. $\endgroup$ Commented May 19, 2013 at 1:34
  • $\begingroup$ What I already have is the solution below (sultan). :) $\endgroup$
    – user78336
    Commented May 19, 2013 at 2:10

1 Answer 1


Substitute $\sqrt[3]{x+2}=u$ therefore $\frac{du}{dx} =\dfrac{1}{3\sqrt[3]{(x+2)^2}}$

Your integral will become : 3$\int\dfrac{u^3(u^3-2)}{u^3+u-2}du $

= $3\int [u^3+ \dfrac{5u-2}{4(u^2+u+2)}-u-\dfrac{1}{4(u-1)}]du$

=$3\int u^3du + \dfrac{3}{4}\int(\dfrac{5(2u+1)}{2(u^2+u+2)}-\dfrac{9}{2(u^2+u+2)})du - \dfrac{3}{4}\int\dfrac{1}{u-1}du -3\int u.du$

Now you can easily see that differentiation of $u^2+u+2 = 2u+1$ so you can substitute $u^2+u+2 = t$ then proceed you will get your result.

In case of further clarification do let me know...

  • $\begingroup$ That first formula for $du$ can't be right. To begin with, it needs to have a $dx$ in it somewhere, but that's not all. $\endgroup$ Commented May 19, 2013 at 2:48
  • $\begingroup$ If $u = \sqrt[3]{x+2}$, then $\frac{du}{dx} = \frac{1}{3\sqrt[3]{(x+2)^2}}$ $\endgroup$
    – daviewales
    Commented May 19, 2013 at 3:59
  • $\begingroup$ Even Wolfram Alpha says so... =P wolframalpha.com/input/… $\endgroup$
    – daviewales
    Commented May 19, 2013 at 4:02
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    $\begingroup$ The edit is better, although the $dx$ is still missing. $\endgroup$ Commented May 19, 2013 at 6:28

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