# Homework Help: Integral of a root polynomial

1. Jul 10, 2009

### Footballer010

1. The problem statement, all variables and given/known data
$$\int\frac{dx}{x^{1/2}-x^{1/3}}$$

2. Relevant equations

None

3. 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?

2. Jul 10, 2009

### Bohrok

Re: Integration

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

3. Jul 10, 2009

### Footballer010

Re: Integration

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?

4. Jul 10, 2009

### diazona

Re: Integration

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

5. Jul 10, 2009

### Footballer010

Re: Integration

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

6. Jul 10, 2009

### Bohrok

Re: Integration

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.

7. Jul 10, 2009

### Footballer010

Re: Integration

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

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

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

Now what?

8. Jul 10, 2009

### Bohrok

Re: Integration

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.

9. Jul 10, 2009

### Footballer010

Re: Integration

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

10. Jul 10, 2009

### diazona

Re: Integration

Did you use Bohrok's suggestion? Solve for dx in terms of u and du and substitute it in to 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. Jul 10, 2009

### Footballer010

Re: Integration

I just got it...

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

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

12. Jul 11, 2009

### Дьявол

Re: Integration

You did good on the start.
$$\int\frac{dx}{u^{3}-u^{2}}$$ 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