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Math Help - please help, desperate!

  1. #1
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    please help, desperate!

    <br /> <br />
\int_{1}^4 3/(2x + 1)^2\, dx<br />
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    Quote Originally Posted by foodmmm View Post
    <br /> <br />
\int_{1}^4 3/(2x + 1)^2\, dx<br />
    Let u=2x+1\Rightarrow du=2\,dx. Substitute.
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  3. #3
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    Pull out the 3 (because it's a constant and you can do that ), and then try u-substitution with u being (2x + 1)
    Last edited by mr fantastic; April 13th 2009 at 08:11 PM. Reason: No edit - flagging post as having been moved from a thread with duplicate question
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  4. #4
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    Quote Originally Posted by foodmmm View Post
    <br /> <br />
\int_{1}^4 3/(2x + 1)^2\, dx<br />
    Notice that this is
    \int_1^4{3(2x + 1)^{-2}\,dx}

     = 3\int_1^4{(2x + 1)^{-2}\,dx}

     = \frac{3}{2}\int_1^4{2(2x + 1)^{-2}\,dx}.

    You would solve this using a u-substitution.

    Let u = 2x + 1 so that \frac{du}{dx} = 2


    So the integral becomes

    \frac{3}{2}\int_{x = 1}^{x = 4}{u^{-2}\,\frac{du}{dx}\,dx}

     = \frac{3}{2}\int_{x = 1}^{x = 4}{u^{-2}\,du}

     = \frac{3}{2}\left[-u^{-1}\right]_{x = 1}^{x = 4}

     = \frac{3}{2}\left[-\frac{1}{2x + 1}\right]_1^4

     = \frac{3}{2}\left[\left(-\frac{1}{9}\right) - \left(-\frac{1}{3}\right)\right]

     = \frac{3}{2}\left(\frac{1}{3} - \frac{1}{9}\right)

     = \frac{3}{2}\left(\frac{2}{9}\right)

     = \frac{1}{3}.
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  5. #5
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    thank you. is there any way to do it without substitution? i don't think we are supposed to use it.
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  6. #6
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    Quote Originally Posted by foodmmm View Post
    thank you. is there any way to do it without substitution? i don't think we are supposed to use it.
    Yes. Know that \int \! (ax + b)^n \, dx = \frac{1}{a} \cdot \frac{1}{n+1} (ax + b)^{n+1}, n \neq -1 (where I have omitted the arbitrary constant).
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