How Do You Integrate 1/(x^(1/2) - x^(1/3))?

Footballer010
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Homework Statement


\int\frac{dx}{x^{1/2}-x^{1/3}}


Homework Equations



None

The Attempt at a Solution



What I did was replace \int\frac{dx}{x^{1/2}-x^{1/3}} with \int\frac{dx}{u^{3}-u^{2}} if u=x^{1/6}. Just to simplify things. And I think that was pointless. Help me out?
 
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I would suggest that you continue with u = x1/6. What was the last thing you did?
 


I used integration by parts and got \frac{1}{u^{2}(u-1)}=\frac{-1}{u}-\frac{1}{u^{2}}+\frac{1}{u-1}. Is this right?
 


Well, first you need to convert the dx to an expression involving du. To do so, start by taking the derivative of u = x^{1/6}.
 


So du=\frac{x^{-5/6}}{6}. Now I'm lost again.
 


Since you don't have x-5/6 (ignoring the constant 1/6) in the integrand, use u = x1/6 to substitute into du = x-5/6/6 dx (don't forget the dx!) so that you only have the variable u in the integrand.
 


u=x^{1/6} so u^{6}=x

du=((u^{6})^{-5/6})/6 dx

So, du=(u^{-5})/6 dx

Now what?
 


You want to get dx by itself, then substitute what it's equal to into your integral. After that you shouldn't have any trouble with it.
 


Hmm...I entered the integral into my calc and its not even close to what I got. This makes no sense!
 
  • #10


Did you use Bohrok's suggestion? Solve for dx in terms of u and du and substitute it into the integral, so that the integral is entirely in terms of u and du. In other words, what is f(u) in this expression?
\int\frac{\mathrm{d}x}{x^{1/2}-x^{1/3}} = \int f(u) \mathrm{d}u
 
  • #11


I just got it...

6\intu^{5}/(u^{2}(u-1))du.

Then I simplified and used long division to get a long complicated answer, but its right. Thank you both!
 
  • #12


You did good on the start.
<br /> \int\frac{dx}{u^{3}-u^{2}}<br /> where u=x^{1/6}

The next step is:

\int\frac{dx}{u^2(u-1)}

What would you do next?

Partial fractions. Then substitution, and finally substitution for x=u^6
 
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