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Indefinite integral

  • Thread starter fluxions22
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  • #1
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Homework Statement


integral of 1/x^2/3(1+x^1/3)


Homework Equations


integral of 1/x dx = ln|x| + c


The Attempt at a Solution



let u= x ^2/3(1+x^1/3)

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
10
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note: the stated relevant equation may actually be 1/u du = ln |u| + C not sure
 
  • #3
33,631
5,288

Homework Statement


integral of 1/x^2/3(1+x^1/3)
This is very ambiguous. What exactly is the integrand?

Homework Equations


integral of 1/x dx = ln|x| + c


The Attempt at a Solution



let u= x ^2/3(1+x^1/3)

Homework Statement





Homework Equations





The Attempt at a Solution

 
  • #4
10
0
the problem is 1 divided by x^2/3(1+x^1/3) dx

the integral is indefinite
 
  • #5
10
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i dont think it calls for u substitution on further review
 
  • #6
33,631
5,288
the problem is 1 divided by x^2/3(1+x^1/3) dx
This is still very ambiguous.
This is what you wrote, but not what I think you meant.
[tex]\frac{1}{\frac{x^2 (1 + \frac{x^1}{3})}{3}}[/tex]
 
  • #7
10
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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
33,631
5,288
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:
[tex]\int \frac{1}{x^{2/3}(1 + x^{1/3})} dx[/tex]

I would start with an ordinary substitution, u = x1/3. I doubt very much that this will turn into du/u.
 
  • #9
139
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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:
[tex]\int \frac{1}{x^{2/3}(1 + x^{1/3})} dx[/tex]

I would start with an ordinary substitution, u = x1/3. I doubt very much that this will turn into du/u.
Then try u = 1+ x1/3, that should do it. I don't think that's too much help, is it?
 
  • #10
10
0
if u is 1+ x^1/3 then ln |1+x^1/3| + C is correct?
 
  • #11
33,631
5,288
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 [itex]\neq[/itex] x^(1/3). The first is the same as x/3. When you have fractional exponents USE PARENTHESES!
 

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