Ask Uncle Colin: A multi-cubic integral

Dear Uncle Colin,

I need to calculate $\int x^3 (x^3+1) (x^3 + 2)^{\frac 13} \dx$ and it's giving me a headache! Can you help?

I've Blundered Using Parts, Rolled Out Fourier Expansions... Nothing!

Hi, IBUPROFEN, and thanks for your message! That’s a bit of a brute, but it can be done with a couple of substitutions.

What's ugly?

One of the first questions I ask myself with something like this is: "What’s ugly?" Here, the $(x^3 + 2)^{\frac{1}{3}}$ isn’t very nice, so I pick the substitution $u = x^3 + 2$, which makes $\diff ux= 3x^2$ ; the middle bracket is $u-1$ and I’ve now got:

$\frac{1}{3}\int x (u-1) u^{\frac{1}{3}} \d u$.

... but I need to get rid of that $x$. I can write it as $(u-2)^{\frac{1}{3}}$, though:

$\frac{1}{3} \int (u-1) (u-2)^{\frac{1}{3}} u^{\frac{1}{3}} \d u$

Now, those two cube-root bits can be combined into one:

$\frac{1}{3} \int (u-1) (u^2 - 2u)^{\frac{1}{3}} \d u$

What's ugly now?

Let’s let $v = u^2 - 2u$, so $\diff v u = 2u - 2$ — or $2(u-1)$1.

This becomes $\frac{1}{6} \int v^{\frac{1}{3}} \d v$, which is now pretty straightforward.


This works just fine - but looking at how it's worked through, I suspect using the substitution $u = x^3 + 1$ would have led to a slightly more obvious second step. Both lead to the same answer, of course.

Tidying up

We get $\frac{1}{8} v^{\frac{4}{3}} + C$, or $\frac{1}{8} [u(u - 2)]^{\frac{4}{3}} + C$, or $\frac{1}{8} [x^3 (x+2)^3 ]^{\frac{4}{3}} + C$ or even - bringing the $x^3$ outside of the square bracket : $\frac{1}{8} x^4 (x+2)^{\frac{4}{3}} + C$.

Hope that helps!



Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. See, @realityminus3, I can do it properly. []


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