nonipaify
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Why do we integrate 1/(1-y^2) by first expressing this as 1/(1-y)(1+y) and then using the partial fractions and so on... Why don't we just integrate it as ln(1-y^2)/-2y? I have understood how it is done but not WHY. I was told that the partial fraction method is used to integrate rational functions of the form a/b where a and b are both polynomials and in 1/(1-y^2) a, which is 1 is not a polynomial cuz it has no variables... HELP
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Mr M
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(Original post by nonipaify)
Why do we integrate 1/(1-y^2) by first expressing this as 1/(1-y)(1+y) and then using the partial fractions and so on... Why don't we just integrate it as ln(1-y^2)/-2y? I have understood how it is done but not WHY. I was told that the partial fraction method is used to integrate rational functions of the form a/b where a and b are both polynomials and in 1/(1-y^2) a, which is 1 is not a polynomial cuz it has no variables... HELP
Try differentiating your "answer".
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user2020user
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(Original post by nonipaify)
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Answering your query as to whether or not a constant is a polynomial, check this out! Also, another way to integrate your integrand would be to use a trigonometric or hyperbolic, which I doubt you've yet to encounter, substitution.
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nonipaify
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(Original post by Mr M)
Try differentiating your "answer".


I get a really complication expression:

(4y^2/(1-y^2) + 2ln(1-y^2)) / 4y^2

By using the quotient rule...
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nonipaify
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(Original post by Khallil)
Answering your query as to whether or not a constant is a polynomial, check this out! Also, another way to integrate your integrand would be to use a trigonometric or hyperbolic, which I doubt you've yet to encounter, substitution.
Are you saying the 1 is actually a polynomial and the expression 1/1-y^2 is indeed a rational function?
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Mr M
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(Original post by nonipaify)
I get a really complication expression:

(4y^2/(1-y^2) + 2ln(1-y^2)) / 4y^2

By using the quotient rule...
So you have answered your question. The "rule" you were attempting to use doesn't exist.

\displaystyle \int \frac{1}{f(x)} dx \neq \frac{\ln f(x)}{f'(x)} + k
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user2020user
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(Original post by nonipaify)
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Yea ... Also, if you don't end up with your integrand (the function you're integrating) when differentiating your final result, you've gone wrong somewhere!
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nonipaify
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#8
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Thank you all for the help. Problem solved.
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