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A level Maths - Integration
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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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#2
(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
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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#3
(Original post by nonipaify)
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(Original post by Mr M)
Try differentiating your "answer".
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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(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.
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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#6
(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...
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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#7
(Original post by nonipaify)
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