Evaluating an indefinite integral

Click For Summary

Homework Help Overview

The discussion revolves around evaluating an indefinite integral involving a combination of terms, including a root and rational expressions. The subject area pertains to integral calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find the anti-derivative of the term involving the fifth root of x raised to the fifth power and expresses uncertainty about the appropriate formula. Some participants suggest simplifying the expression before integrating, while others clarify the application of exponent rules.

Discussion Status

Participants are actively engaging in the problem, with some providing guidance on simplification and the correct approach to finding the anti-derivative. There is an acknowledgment of the need for a single constant of integration for the entire expression.

Contextual Notes

There is a mention of the original poster's difficulty in finding relevant formulas in their textbook, indicating potential gaps in their understanding of the material.

Wm_Davies
Messages
51
Reaction score
0

Homework Statement


Evaluate the indefinite integral.

\int \left({\sqrt[5]{x^5}}-\frac{6}{5 x}+\frac{1}{4 x^{7}} \right) dx


The Attempt at a Solution



O.k. the only anti-derivative I am having trouble getting is the first one {\sqrt[5]{x^5}}.

I am not sure what formula I would use or how to do it. I looked through the book, but I didn't see anything addressing this. Any help would be appreciated.

I imagine that it would be easier to write it as (x^{5})^{\frac{1}{5}}

Working through it I get (\frac{x^{6}}{6})^{\frac{1}{5}}

then I am stuck...
 
Physics news on Phys.org
Remember that when you raise a power to a power, that by the Law of Exponents the powers multiply. :wink:
 
Simplify first!
\sqrt[5]{x^5}~=~x
 
Mark44 said:
Simplify first!
\sqrt[5]{x^5}~=~x

Yeah, I can't believe I missed that. So then the anti-derivative should be \frac{x^{2}}{2} + a constant?
 
Yes, but you won't need a constant for each term in the integrand - just one for all three.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
2K
Replies
4
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 27 ·
Replies
27
Views
3K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K