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Thread: Quick integration question

  1. #1
    Super Member craig's Avatar
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    Quick integration question

    Integrate wrt x, $\displaystyle \frac{1}{4x^2 + 4x +17}dx$.

    I rearranged it into the form $\displaystyle \frac{1}{(2x+1)^2 + 16}dx$.

    I know this integrates to some form of arctan, and I though that it would be

    $\displaystyle \frac{1}{4} arctan{(\frac{2x+1}{4})}$.

    But in the book they have the answer as $\displaystyle \frac{1}{8} arctan{(\frac{2x+1}{4})}$.

    Is this because you need to divide by the derivative of $\displaystyle (2x+1)$ as well, or is there something I'm missing?

    Thanks
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  2. #2
    Super Member craig's Avatar
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    Oh and I know I'm missing the constant, but this is part of a definite integral question.
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  3. #3
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    Let y = 2x+1 then dy = 2 dx

    $\displaystyle \int \frac{dx}{(2x+1)^2+16} = \frac12 \:\int \frac{dy}{y^2+16}$

    Let y = 4z then dy = 4 dz

    $\displaystyle \int \frac{dx}{(2x+1)^2+16} = \frac12 \: \int \frac{4 dz}{16(z^2+1)} = \frac18 \int \frac{dz}{z^2+1}$

    $\displaystyle \int \frac{dx}{(2x+1)^2+16} = \frac18 \:\arctan(z) + c = \frac18 \:\arctan\left(\frac{2x+1}{4}\right) + c$
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  4. #4
    Moo
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    Hello,
    Quote Originally Posted by craig View Post
    Integrate wrt x, $\displaystyle \frac{1}{4x^2 + 4x +17}dx$.

    I rearranged it into the form $\displaystyle \frac{1}{(2x+1)^2 + 16}dx$.

    I know this integrates to some form of arctan, and I though that it would be

    $\displaystyle \frac{1}{4} arctan{(\frac{2x+1}{4})}$.

    But in the book they have the answer as $\displaystyle \frac{1}{8} arctan{(\frac{2x+1}{4})}$.

    Is this because you need to divide by the derivative of $\displaystyle (2x+1)$ as well, or is there something I'm missing?

    Thanks
    It would be 1/4 if it were $\displaystyle \frac{1}{(x+\dots)^2+16}$
    But here, it's as if you omit the chain rule, since there's a coefficient in front of x.
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  5. #5
    Super Member craig's Avatar
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    Thanks for both replies.

    Quote Originally Posted by Moo View Post
    It would be 1/4 if it were $\displaystyle \frac{1}{(x+\dots)^2+16}$
    But here, it's as if you omit the chain rule, since there's a coefficient in front of x.
    So if for example I had $\displaystyle \int \frac{1}{(3x+1)^2 + 16}dx$, the answer would be $\displaystyle \frac{1}{12} \:\arctan\left(\frac{3x+1}{4}\right) + c$

    Thanks again
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  6. #6
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    Quote Originally Posted by craig View Post
    Thanks for both replies.



    So if for example I had $\displaystyle \int \frac{1}{(3x+1)^2 + 16}dx$, the answer would be $\displaystyle \frac{1}{12} \:\arctan\left(\frac{3x+1}{4}\right) + c$

    Thanks again
    Yes. To see that more explicitely, use the substitution u= 3x+1.
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  7. #7
    Super Member craig's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Yes. To see that more explicitely, use the substitution u= 3x+1.
    Thanks
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