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## Homework Statement

[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.)

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- Thread starter mllamontagne
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[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.)

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SammyS

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Hello mllamontagne. Welcome to PF !## Homework Statement

[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.)

## 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.

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

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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.

- #4

HallsofIvy

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You know that [itex]x^3- 1= (x- 1)(x^2+ x+ 1)[/itex] don't you?

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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

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SammyS

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

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

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