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    Someone please help, I'm having a mind block, how do i integrate 1/x(x^2+1)
    Thankyou!
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    Partial fractions?

    \frac{1}{x(x^2+1)}=\frac{A}{x} + \frac{Bx+c}{x^2+1}
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    (Original post by G07153)
    Someone please help, I'm having a mind block, how do i integrate 1/x(x^2+1)
    Thankyou!
    Expand that brackets first to get x^3 + x

    Then write it as (x^3 + x)^-1

    Then intergrate as normal
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    (Original post by G07153)
    Someone please help, I'm having a mind block, how do i integrate 1/x(x^2+1)
    Thankyou!
    If you mean 1/(x(x^2 + 1))dx, then that's the same as (x^3 + x)^-1 dx.

    What happens when you integrate something to the power of minus 1?

    Or make partial fractions, of course
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    I need to use the partial fractions method by the looks of the solutions, but haven't used partial fractions for SO long, could someone run through the method? Thanks for all your help!
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    (Original post by Mr M)
    Partial fractions?

    \frac{1}{x(x^2+1)}=\frac{A}{x} + \frac{Bx+c}{x^2+1}
    How do I go about finding A, B & C??
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    (Original post by G07153)
    How do I go about finding A, B & C??
    Multiply through by x and then set x = 0. You now have A.

    Pick two other values for x (e.g. x = 1 and x = 2) and substitute these to get a pair of simultaneous equations for B and C.
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    (Original post by TGH1)
    What happens when you integrate something to the power of minus 1?
    You get Zorro.
    Spoiler:
    Show

    Wait.. was is zero?
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    (Original post by fletchdd02)
    Then write it as (x^3 + x)^-1

    Then intergrate as normal
    (Original post by TGH1)
    If you mean 1/(x(x^2 + 1))dx, then that's the same as (x^3 + x)^-1 dx.

    What happens when you integrate something to the power of minus 1?
    I hope you two aren't suggesting what I think you're suggesting... there are two separate reasons why this doesn't work - do you know what they are?
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    (Original post by nuodai)
    I hope you two aren't suggesting what I think you're suggesting... there are two separate reasons why this doesn't work - do you know what they are?
    what are they?
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    (Original post by G07153)
    How do I go about finding A, B & C??
    Turn the right hand side into a single fraction over the common denominator x(1+x^2). You do this by cross multiplication.

    Therefore when you equate numerators on each side you would have.

    1 = (A+B)x^2 + Cx + A

    Now compare the coefficients of the powers of x on each side. There is no x^2 term on the left hand side so A+B = 0. Similarly there is no x, so C=0. Finally comparing coefficients of the constants you have 1=A => A=1.

    Therefore A+B=0 => 1+B=0 => B=-1.

    So the the fraction in partials is 1/x - x/(1+x^2), which you can easily integrate. The numerators are the derivatives (up to a constant) of the denominators.
    Remember integral (f'(x)/f(x)) = ln|f(x)|.
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    (Original post by fletchdd02)
    what are they?
    Maybe just one reason, though it depends on what method you intended. The big one is this: whilst it is the case that \dfrac{d}{dx}(f(g(x))=f'(g(x)) \times g'(x) (chain rule) it is not the case that \displaystyle \int f'(g(x))\, dx = \dfrac{f(g(x))}{g'(x)} + C.

    My guess is that you were advising the OP to do \displaystyle \int \dfrac{1}{x^3+x}\, dx = \dfrac{\ln \left| x^3+x \right|}{3x^2+1} + C. This isn't the case (differentiate the RHS to check).

    This is only the case when g(x) is a linear polynomial; for instance it is true that \displaystyle \int \dfrac{1}{3x+2}\, dx = \dfrac{\ln \left| 3x+2 \right|}{3} + C.

    In general if you want to reverse the chain rule then you need to use integration by substitution; you can't just divide by the derivative.

    The second reason is to do with the way you phrased your post: I was worried you were suggesting changing the power to -2 and dividing or whatever. This is wrong because with integration the power is meant to increase, e.g. \displaystyle \int x^{-2}\, dx = -x^{-1}+C. But you can't do this with -1 because then you get 0 and you can't divide by 0. I'm not sure if this is actually what you had in your head, but I thought I'd mention it for completeness.
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    (Original post by nuodai)
    ...
    I'm fairly sure they were suggesting add one to the power and divide by the new power. You just saved them from an abominable and certain death.
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    (Original post by G07153)
    Someone please help, I'm having a mind block, how do i integrate 1/x(x^2+1)
    Thankyou!
    (x power 2 +1) integrate 1/x + whole integrate(derivative of (x power 2 +1) integration of 1/x
    now
    (x power 2 +1)(natural log of x) + whole integration (2x mutiply natural log of x)
    repeat again you will get the solution.
    For free online Education and free physics notes check this http://www.notes4all.com
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    (Original post by mahnoorbloch)
    (x power 2 +1) integrate 1/x + whole integrate(derivative of (x power 2 +1) integration of 1/x
    now
    (x power 2 +1)(natural log of x) + whole integration (2x mutiply natural log of x)
    repeat again you will get the solution.
    You are going to be a barrel of laughs. The spam advertising doesn't help by the way.
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    (Original post by Mr M)
    I'm fairly sure they were suggesting add one to the power and divide by the new power. You just saved them from an abominable and certain death.
    Ahahaha
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    (Original post by fletchdd02)
    Expand that brackets first to get x^3 + x

    Then write it as (x^3 + x)^-1

    Then intergrate as normal
    Would anyone mind doing a step by step method to show how you would integrate this? Sorry but I don't really understand some of the methods posted, I've just finished Core 1 and Core 2 in OCR (I don't recall that a/x method or partial fractions etc in my lessons & they're not in any of my books). I would have originally done what fletchdd did but upon realising that you get zero [and of course you can't divide by zero!] i was shocked! :eek: :rolleyes:
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    (Original post by sach21sk)
    Would anyone mind doing a step by step method to show how you would integrate this? Sorry but I don't really understand some of the methods posted, I've just finished Core 1 and Core 2 in OCR (I don't recall that a/x method or partial fractions etc in my lessons & they're not in any of my books). I would have originally done what fletchdd did but upon realising that you get zero [and of course you can't divide by zero!] i was shocked! :eek: :rolleyes:
    This is A2 maths not AS.
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    (Original post by Mr M)
    This is A2 maths not AS.
    Haha, phew
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    (Original post by sach21sk)
    Haha, phew
    Actually, from the thread title, it is undergraduate maths (although most A2 students would probably know how to approach it if prompted).
 
 
 
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