Trouble with the integral Help

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The integral discussed is \(\int \frac{x^3-1}{4x^3+x} \, dx\). The correct solution involves performing polynomial long division due to the degree of the numerator being greater than or equal to that of the denominator. The final result is \(\frac{x}{4} + \frac{x}{16} \ln \left(\frac{x^{16}}{(2x-1)^7(2x+1)^9}\right)\). Key techniques include factoring the numerator and denominator, specifically recognizing \(x^3 - 1\) as a difference of cubes and applying partial fraction decomposition.

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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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