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Math Help - Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

  1. #1
    Super Member Matt Westwood's Avatar
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    Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

    I'm working my way through a series of ever tougher integrals. I'm stuck at no. 283:

    \int \frac {dx} {x \sqrt {a x^2 + b x + c} }

    ... and there are some further even tougher ones.

    I understand that it is supposed to evaluate to:

    - \frac 1 {\sqrt c} \ln \left({\frac {2 \sqrt c \sqrt {a x^2 + b x + c} + bx + 2c} x} }\right)

    or:

    - \frac 1 {\sqrt c} \sinh^{-1} \left({ \frac {bx + 2 c} {|x| \sqrt {4 a c - b^2} } }\right)

    depending on the sign of b^2 - 4 a c

    I have already tried:
    a) completing the square on the expression under the square root
    b) substituting z = 2 a x + b and z = (2 a x + b)^2
    c) integrating by parts with u = \sqrt {a x^2 + b x + c} and dv = x etc.

    I've also tried working backward, differentiating the \sinh^{-1} expression, to see where it gets me, but the penny is not dropping.

    Any hints?

    The good cause this is for is ProofWiki which now has over 10000 proofs up.

    Thanks, guys.
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  2. #2
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    Re: Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

    It's not elegant but ...

    x \sqrt {a x^2 + b x + c}=\frac{x}{(a x^2 + b x + c)^{-\frac 12}}=\frac{x[2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}{(a x^2 + b x + c)^{-\frac 12} [2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}=

    \frac{x[2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}{[2 \sqrt c +(bx+2c) (ax^2+bx+c)^{-\frac 12}]}=\frac{num}{den}

    where:

    num=\frac{[2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}{x}

    den=\frac{[2 \sqrt c +(bx+2c) (ax^2+bx+c)^{-\frac 12}]}{x^2}

    differentiating num w.r.t. x gives:

    num'=\frac{x[\sqrt c (2ax+b)(ax^2+bx+c)^{-\frac 12}+b]-[2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}{x^2}

    num'=(ax^2+bx+c)^{-\frac 12} \frac{[\sqrt c (2ax^2+bx)+bx \sqrt{ax^2+bx+c}]-[2 \sqrt c (ax^2+bx+c)+(bx+2c)\sqrt{ax^2+bx+c}])}{x^2}

    num'=(ax^2+bx+c)^{-\frac 12} \frac{[\sqrt c (2ax^2+bx -2ax^2 -2bx -2c)+bx \sqrt{ax^2+bx+c}]-[(bx+2c)\sqrt{ax^2+bx+c}])}{x^2}

    num'=(ax^2+bx+c)^{-\frac 12} \frac{[\sqrt c (-bx -2c)-[(2c)\sqrt{ax^2+bx+c}])}{x^2}

    num'=-\frac{2c +\sqrt c (bx +2c)(ax^2+bx+c)^{-\frac 12}}{x^2}=-\sqrt c \times den

    Now provided num is real and non zero the original integral is

     \int \frac{den}{num}dx=\frac {-1}{\sqrt c}\int \frac {num'}{num}=\frac {-1}{\sqrt c} ln(num)=\frac {-1}{\sqrt c} ln \left(\frac{[2 \sqrt c \sqrt{ax^2+bx+c}+bx+2c]}{x} \right)
    Last edited by Kiwi_Dave; July 15th 2014 at 10:41 PM.
    Thanks from Matt Westwood and topsquark
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  3. #3
    Super Member Matt Westwood's Avatar
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    Re: Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

    That works for me. Many thanks.
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  4. #4
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    Re: Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

    I've not followed it all the way through, but the substitution x = 1/u , followed by a completing of the square and then a second substitution seems to work well.
    Thanks from Matt Westwood
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  5. #5
    Super Member Matt Westwood's Avatar
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    Re: Integral of 1 / (x sqrt (ax^2 + b x + c)) dx

    Quote Originally Posted by BobP View Post
    I've not followed it all the way through, but the substitution x = 1/u , followed by a completing of the square and then a second substitution seems to work well.
    I'm just about to try it now -- thank you for this, it seems to clear the x from the denominator leaving one with the integral of \frac 1 {\sqrt {a + bx + c x^2} } whose solution I have already evaluated.
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