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Math Help - check my substitution to find indefinate integral

  1. #1
    Junior Member shepherdm1270's Avatar
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    check my substitution to find indefinate integral

    Integral of x^2+2 / x^3 + 6x + 3

    u = x^3+6x + 3
    du = 3x^2 + 6 = (1/2) du

    (1/2) integral of du/u = (1/2) ln|u| + c

    = (1/2)ln |x^3+6x+3| + C


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  2. #2
    Moo
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    P(I'm here)=1/3, P(I'm there)=t+1/3
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    Hello,

    Where does 1/2 come from ?

    u = x^3+6x + 3 --> ok
    du/dx = 3x^2 + 6 = 3(x^2+2) and this is the numerator.

    So the integral is now \int \frac{1}{3} \frac{du}{u}
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  3. #3
    Junior Member shepherdm1270's Avatar
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    oh right, stupid mistake


    so it would simply be 1/3 in place of the 1/2 and the rest is correct?
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  4. #4
    Moo
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    It seems that... yes ^^
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