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

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

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

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

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

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

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

    Thanks
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    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

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

    Let y = 4z then dy = 4 dz

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

    \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, \frac{1}{4x^2 + 4x +17}dx.

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

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

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

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

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

    Thanks
    It would be 1/4 if it were \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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    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 \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 \int \frac{1}{(3x+1)^2 + 16}dx, the answer would be \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 \int \frac{1}{(3x+1)^2 + 16}dx, the answer would be \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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