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Math Help - Integral ( complete the square )

  1. #1
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    Integral ( complete the square )

    I have a problem in my book where i have an integral and i have to use complete the square if necessary. I understand how to do the complete the square. However, this problem don't need it but its stumping me. I am not sure how the solutions manual got the second step. It split it up into two integrals and i cant see how it came to that conclusion?

     \int \frac {2x}{x^2+6x+13} dx

    The next step is

        \int \frac {2x + 6} {x^2 + 6x + 13} dx - 6 \int \frac {1}{x^2+6x+13} dx

    How are they splitting up the integral and where is the 2x +6 in the first integral and -6 in the second coming from?

    The next step( which i left out) looks like they use complete the square on the second integral.

    Thank you so much!!!!!
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  2. #2
    Super Member Quacky's Avatar
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    Re: Integral ( complete the square )

    \int~\dfrac{2x}{x^2+6x+13} dx

    =\int~\dfrac{2x+(6-6)}{x^2+6x+13} dx

    =\int~\dfrac{2x+6}{x^2+6x+13}dx-\int~\dfrac{6}{x^2+6x+13} dx

    =\int~\dfrac{2x+6}{x^2+6x+13}dx-6\int~\dfrac{1}{x^2+6x+13} dx

    Then, I think you probably would complete the square on the second integral.
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  3. #3
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    Re: Integral ( complete the square )

    Why did they choose the 6 though? Why couldn't i have used any number?
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  4. #4
    Super Member Quacky's Avatar
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    Re: Integral ( complete the square )

    Because using the 6 makes the first integral easy to solve. This is because it is now of the form:

    \int~\dfrac{f'(x)}{f(x)} dx = ln(f(x))+ C
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  5. #5
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    Re: Integral ( complete the square )

    I understand your formula. However, i am still confused on how i would have known to use a 6? You said it make it easier but how would i have known that? Why couldnt i have used a 3 instead? Im sorry but im just baffled.
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  6. #6
    Super Member Quacky's Avatar
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    Re: Integral ( complete the square )

    I understand your confusion. When you see a fraction that you have to integrate, there are a few things to try. It really is a case of "manipulate, guess and check."

    -Partial fractions. Here this isn't possible because the denominator doesn't factor.
    -Substitutions. Here, there *might* be a substitution that will make life easier but I certainly wouldn't spot it easily.
    If neither of these will work, then check whether you can somehow make something of the form \frac{f'(x)}{f(x)} - what my teacher used to call a "related integral" for some reason.

    Start with the denominator. Here, this is x^2+6x+13. If we differentiate this, we get 2x+6. If this is our numerator, then great! We have something of the form \frac{f'(x)}{f(x)}. But it isn't.

    \frac{2x+6}{x^2+6x+13} is of the form \frac{f'(x)}{f(x)} but \frac{2x}{x^2+6x+13} is not.

    We cannot easily integrate

    \frac{2x}{x^2+6x+13}

    ...so we make it into:

    \frac{2x+6}{x^2+6x+13}-\frac{6}{x^2+6x+13} which we can then integrate. We don't try '3' because this does not relate to the denominator. We get the '6' from differentiating the denominator.
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  7. #7
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    Re: Integral ( complete the square )

    Perfect! That explained it perfectly. The U prime over U should have been a clue to me. Thank you so much!!!!
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