1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Partial fractions for a cubic root in the denominator of integrand

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]\int\frac{1}{x\sqrt[3]{x+1}}[/itex]dx (That's a cubic root in the denominator, by the way. Not an x cubed.)

    3. The attempt at a solution I thought possibly partial fractions, but I've never seen it done with a root in the denominator. Integration by parts was unsuccessful. Thanks so much if anyone can show me the steps for this. Must be able to broken up into several bits that integrate to logs or trig functions or something, but I can't seem to get it.
  2. jcsd
  3. Sep 3, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Hello mllamontagne. Welcome to PF !

    Try a substitution instead: perhaps [itex]u=\sqrt[3]{x+1}\ .[/itex]
  4. Sep 4, 2012 #3
    Thanks for the welcome and the advice.
    If I u substitute with u=[itex]\sqrt[3]{x+1}[/itex], I get
    3∫[itex]\frac{u}{u^{3}-1}[/itex]du which admittedly looks better, but which I still can't figure out how to solve.
  5. Sep 4, 2012 #4


    User Avatar
    Science Advisor

    The method of "partial fractions" typically only works for rational functions- fractions with polynomials on top and bottom. Once you have made the substitution SammyS suggested, that is exactly what you have and you can use "partial fractions".
    You know that [itex]x^3- 1= (x- 1)(x^2+ x+ 1)[/itex] don't you?
  6. Sep 4, 2012 #5
    As you suggest, Partial fractions can be used to break up

    3∫[itex]\frac{u}{(u-1)(u^{2}+u+1)}[/itex]du with u=[itex]\sqrt[3]{x+1}[/itex]

    to ∫[itex]\frac{1}{u-1}[/itex]du +∫[itex]\frac{1}{u^{2}+u+1}[/itex] du -∫[itex]\frac{u}{u^{2}+u+1}[/itex] du

    the first integral is a simple natural log, the second is and arctan, but now the third one gives me trouble. how would I accomplish the operation below?

    ∫[itex]\frac{u}{u^{2}+u+1}[/itex] du
  7. Sep 4, 2012 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Split up [itex]\displaystyle \frac{1-u}{u^{2}+u+1}[/itex] a bit differently.

    If you have [itex]\displaystyle \frac{2u+1}{u^{2}+u+1}[/itex] then the numerator is the derivative of the denominator & you will have an anti-derivative which is a logarithm.

    For the numerator, notice that

    [itex]\displaystyle 1-u[/itex]

    [itex]\displaystyle =1-\frac{1}{2}\left(2u+1\right)+\frac{1}{2}[/itex]​
  8. Sep 4, 2012 #7
    SammyS, you are the best. That was exactly what I needed to do. I promise I will never forget how to do that integration again.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook