Trouble with the integral Help

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Homework Help Overview

The discussion revolves around solving the integral \(\int \frac{x^3-1}{4x^3+x}\). The original poster expresses difficulty in obtaining the correct result as provided in their textbook.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • One participant suggests performing polynomial long division due to the degree of the numerator being greater than or equal to that of the denominator. Another participant notes the numerator can be factored as a difference of cubes and mentions the possibility of using partial fractions for the denominator.

Discussion Status

Participants are exploring different approaches to tackle the integral, including polynomial long division and factoring. There is no explicit consensus yet, but suggestions for methods have been provided.

Contextual Notes

The original poster is working from a textbook example and is seeking clarification on their approach to the integral, indicating a potential gap in understanding the necessary steps to solve it correctly.

depecheSoul
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Trouble with the integral! Help

While I was working some examples of integrals from book, I found this example:

\int {{x^3-1}\over{4x^3+x}}
and I can not solve it, because I get always incorrect result.

Result is: (from book)

{x \over 4}+{{x \over 16}*{ln ({{x^{16}}\over{(2x-1)^7(2x+1)^9}}})}

Can you help me solve it!

Thanks.
 
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Before you do anything with the integral, take polynomial long division, since the degree of the numerator >= degree of the denominator.
 


depecheSoul said:
While I was working some examples of integrals from book, I found this example:

\int {{x^3-1}\over{4x^3+x}}
and I can not solve it, because I get always incorrect result.

Result is: (from book)

{x \over 4}+{{x \over 16}*{ln ({{x^{16}}\over{(2x-1)^7(2x+1)^9}}})}

Can you help me solve it!

Thanks.

You can notice x^3 - 1 is of the form "a^3-b^3". You can factorize ;) !

And you can factorize: 4x^3 + x.

It is a partial fraction.

Good luck ;) !
 


Thank you!
 

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