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