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Math Help - Integrals leading to a Log function

  1. #1
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    Integrals leading to a Log function

    Hi all,
    I am working on a problem at the moment which i have the answer for but i don't understand one step of the solution.

    Evaluate "integration sign" (2x-6)dx/[(x^2)-6x+10]

    The answer is given as Ln2.

    I can see how the answer comes about but a step of it is confusing me.

    Once i substitue and solve i get Ln[1x^2)-6(x)+10]/??? as per the rule for integrating 1/ax+b=ln(ax+b)/a

    In otherwords 1/ax+b=(1[ln(ax+b)]/a) +c

    However which value do i pic for a? It seems there are 2 coefficients of x in this case, 1 for the x squared, and -6 for the x.

    It looks like the a chosen in the solution is the coefficient of the X^2 which is 1.

    Why is the coefficient of the -6x not used for the a ?

    It seems a pretty fundemental issue that i need solved.

    When given a quadratic expression & when integrating is the a coefficient for the solution always the coefficient associated with the x^2?

    Apologies if this is a ridiculous question
    John
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Integrals leading to a Log function

    You're right that:
    \int \frac{2x-6}{x^2-6x+10}dx= \ln|x^2-6x+10|+C
    I think it's strange there're no integration limits given.
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  3. #3
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    Re: Integrals leading to a Log function

    I think we have to take t = [(x^2)-6x+10]. Then dt = 2x-6. If you substitute this, your problem will become integration of dt/t.
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  4. #4
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    Re: Integrals leading to a Log function

    sorry guys the limits are b=4 a=3
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Integrals leading to a Log function

    That changes the cases, that means we have:
    [\ln|x^2-6x+10|]_{3}^{4}=\ln|4^2-24+10|-\ln|3^2-18+10|=\ln|2|-\ln|1|=\ln|2|-0=\ln(2)
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  6. #6
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    Re: Integrals leading to a Log function

    i should have just solved the expression -tut!
    thanks all
    John
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  7. #7
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    Re: Integrals leading to a Log function

    But if you are going to make the substitution t= x^2- 6x+ 10, dx= (2x- 6)dx, you might as well change the limits of integration also: when x= 3, t= 9- 18+ 10= 1 and when x= 4 t= 16- 24+ 10= 2 so
    \int_3^4 \frac{2x- 6}{x^2- 6x+ 10}dx= \int_1^2\frac{dt}{t}= \left[ln(t)\right]_1^2= ln(2)- ln(1)= ln(2).
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