• Support PF! Buy your school textbooks, materials and every day products Here!

Partial fractions for a cubic root in the denominator of integrand

  • #1

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.
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,304
998

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.
Hello mllamontagne. Welcome to PF !

Try a substitution instead: perhaps [itex]u=\sqrt[3]{x+1}\ .[/itex]
 
  • #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.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
955
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?
 
  • #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
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,304
998
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.
 
  • #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.
 

Related Threads on Partial fractions for a cubic root in the denominator of integrand

Replies
3
Views
2K
Replies
9
Views
2K
Replies
3
Views
2K
Replies
2
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
1
Views
995
Top