Partial fractions for a cubic root in the denominator of integrand

mllamontagne · Sep 3, 2012

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.

SammyS · Sep 3, 2012

mllamontagne said: ↑

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.

Hello mllamontagne. Welcome to PF !

Try a substitution instead: perhaps [itex]u=\sqrt[3]{x+1}\ .[/itex]

mllamontagne · Sep 4, 2012

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.

HallsofIvy · Sep 4, 2012

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?

mllamontagne · Sep 4, 2012

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

SammyS · Sep 4, 2012

mllamontagne said: ↑

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

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+1-1\right)[/itex]

[itex]\displaystyle =1-\frac{1}{2}\left(2u+1\right)+\frac{1}{2}[/itex]
etc.

mllamontagne · Sep 4, 2012

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.

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Partial fractions for a cubic root in the denominator of integrand

mllamontagne

SammyS

mllamontagne

HallsofIvy

mllamontagne

SammyS

mllamontagne

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