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

Click For Summary

Homework Help Overview

The problem involves integrating the function \(\int\frac{dx}{x^{1/2}-x^{1/3}}\), which falls under the subject area of calculus, specifically integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss a substitution \(u = x^{1/6}\) to simplify the integral. There are attempts to express \(dx\) in terms of \(du\) and to manipulate the integrand accordingly. Some participants express confusion about the steps and seek clarification on how to proceed with the integration.

Discussion Status

The discussion is ongoing, with participants providing guidance on the substitution process and expressing varying levels of understanding. Some have successfully transformed the integral, while others are still grappling with the necessary steps to complete the integration.

Contextual Notes

There is mention of a calculator's output not matching participants' results, indicating potential misunderstandings or errors in the integration process. The discussion also highlights the need to ensure all variables are correctly substituted in the integral.

Footballer010
Messages
14
Reaction score
0

Homework Statement


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


Homework Equations



None

The Attempt at a Solution



What I did was replace [tex]\int\frac{dx}{x^{1/2}-x^{1/3}}[/tex] with [tex]\int\frac{dx}{u^{3}-u^{2}}[/tex] if u=x[tex]^{1/6}[/tex]. Just to simplify things. And I think that was pointless. Help me out?
 
Physics news on Phys.org


I would suggest that you continue with u = x1/6. What was the last thing you did?
 


I used integration by parts and got [tex]\frac{1}{u^{2}(u-1)}[/tex]=[tex]\frac{-1}{u}[/tex]-[tex]\frac{1}{u^{2}}[/tex]+[tex]\frac{1}{u-1}[/tex]. 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 [itex]u = x^{1/6}[/itex].
 


So du=[tex]\frac{x^{-5/6}}{6}[/tex]. 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[tex]^{1/6}[/tex] so u[tex]^{6}[/tex]=x

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

So, du=(u[tex]^{-5}[/tex])/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 [itex]f(u)[/itex] in this expression?
[tex]\int\frac{\mathrm{d}x}{x^{1/2}-x^{1/3}} = \int f(u) \mathrm{d}u[/tex]
 
  • #11


I just got it...

6[tex]\int[/tex][tex]u^{5}/(u^{2}(u-1)[/tex])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.
[tex] \int\frac{dx}{u^{3}-u^{2}}[/tex] where [tex]u=x^{1/6}[/tex]

The next step is:

[tex]\int\frac{dx}{u^2(u-1)}[/tex]

What would you do next?

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

Similar threads

  • · Replies 105 ·
4
Replies
105
Views
11K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
7
Views
2K
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K