Indefinite Integral of 1/x^2/3(1+x^1/3) | Homework Equations & Solution

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Homework Help Overview

The discussion revolves around the indefinite integral of the expression 1/x^(2/3)(1+x^(1/3)). Participants are attempting to clarify the integrand and explore potential methods for solving the integral.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the clarity of the integrand and suggesting different interpretations of the expression. There are attempts to define the integrand more clearly, with some suggesting substitutions and others expressing uncertainty about the necessity of u-substitution.

Discussion Status

The discussion is ongoing, with various interpretations of the integrand being explored. Some participants have provided guidance on how to express the integral more clearly, while others are still questioning the setup and assumptions involved.

Contextual Notes

There is ambiguity in the original statement of the problem, leading to confusion about the correct form of the integrand. Participants are also discussing the importance of using parentheses in expressions involving fractional exponents.

fluxions22
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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)
 
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note: the stated relevant equation may actually be 1/u du = ln |u| + C not sure
 
fluxions22 said:

Homework Statement


integral of 1/x^2/3(1+x^1/3)
This is very ambiguous. What exactly is the integrand?
fluxions22 said:

Homework Equations


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


The Attempt at a Solution



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

the integral is indefinite
 
i don't think it calls for u substitution on further review
 
fluxions22 said:
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.
\frac{1}{\frac{x^2 (1 + \frac{x^1}{3})}{3}}
 
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)
 
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.
 
Mark44 said:
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.

Then try u = 1+ x1/3, that should do it. I don't think that's too much help, is it?
 
  • #10
if u is 1+ x^1/3 then ln |1+x^1/3| + C is correct?
 
  • #11
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!
 

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