# Homework Help: Indefinite integral

1. May 6, 2010

### fluxions22

1. The problem statement, all variables and given/known data
integral of 1/x^2/3(1+x^1/3)

2. Relevant equations
integral of 1/x dx = ln|x| + c

3. The attempt at a solution

let u= x ^2/3(1+x^1/3)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 6, 2010

### fluxions22

note: the stated relevant equation may actually be 1/u du = ln |u| + C not sure

3. May 6, 2010

### Staff: Mentor

This is very ambiguous. What exactly is the integrand?

4. May 6, 2010

### fluxions22

the problem is 1 divided by x^2/3(1+x^1/3) dx

the integral is indefinite

5. May 6, 2010

### fluxions22

i dont think it calls for u substitution on further review

6. May 6, 2010

### Staff: Mentor

This is still very ambiguous.
This is what you wrote, but not what I think you meant.
$$\frac{1}{\frac{x^2 (1 + \frac{x^1}{3})}{3}}$$

7. May 6, 2010

### fluxions22

the ^ sign stands for "to the power of" but if i express the problem in words it would be "one divided by x to the 2/3 power times (1 plus x to the one thirds power)

8. May 6, 2010

### Staff: Mentor

Then you should write the integrand as 1/[x^(2/3)(1+x^(1/3))]. Note the parentheses around the exponents.

Better yet, here's the LaTeX for your integral:
$$\int \frac{1}{x^{2/3}(1 + x^{1/3})} dx$$

I would start with an ordinary substitution, u = x1/3. I doubt very much that this will turn into du/u.

9. May 6, 2010

### physicsman2

Then try u = 1+ x1/3, that should do it. I don't think that's too much help, is it?

10. May 8, 2010

### fluxions22

if u is 1+ x^1/3 then ln |1+x^1/3| + C is correct?

11. May 8, 2010

### Staff: Mentor

Not quite. If u = 1 + x^(1/3), du = dx/(3x^(2/3)). Your answer needs to account for that factor of 3 in the denominator.

Note that x^1/3 $\neq$ x^(1/3). The first is the same as x/3. When you have fractional exponents USE PARENTHESES!