# Improper Fraction Integral (Partial Fraction Probably?)

• Jun 1st 2012, 09:38 AM
jphillips
Improper Fraction Integral (Partial Fraction Probably?)
Hi. I've been reviewing for Calculus III coming up this summer and this problem really has be stumped:

http://www4c.wolframalpha.com/Calcul...s=59&w=54&h=51

I've broken it up into x1/3/x1/3-1 + 1/x1/3-1 and then u-sub u=x1/3, but now am confused about the next step.
• Jun 1st 2012, 10:01 AM
HallsofIvy
Re: Improper Fraction Integral (Partial Fraction Probably?)
Well, you separated it to " $x^{1/3}/(x^{1/3}- 1)+ 1/(x^{1/3}- 1)$" which is not quite what you wrote!

Rather that doing it that way, I think you would be better off actually dividing to get $\frac{x^{1/3}+ 1}{x^{1/3}- 1}= 1+ \frac{2}{x^{1/3}- 1}$. Once you have done that, get rid of the entire denominator in the second part by the substitution $u= x^{1/3}- 1$. Then, of course, $du= (1/3)x^{-2/3}dx$ so that $dx= 3x^{2/3}du$. Since $x^{1/3}= u+ 1$ so that $x^{2/3}= (u+ 1)^2$.
• Jun 4th 2012, 06:02 AM
jphillips
Re: Improper Fraction Integral (Partial Fraction Probably?)
Ah! That's tricky! So when you substitute you will end up with 6((u+1)^2)du/u and can foil then power rule. I think I remember learning the rule that when dividing out (your first step) the power in the numerator must be greater than or equal to the power in the denominator? (Just trying to establish a way that I can recognize when to do this.)

Thanks much!