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Math Help - integration

  1. #1
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    integration

    How do you integrate 1/(2v + 1)^2?
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  2. #2
    Senior Member yeKciM's Avatar
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    2v+1 = u
    du=2dv


    and so on ...

    u must get :

     \displaystyle \int \frac {1}{(2v+1)^2} \;dv= -\frac {1}{4v+2}+ C
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  3. #3
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    Quote Originally Posted by CookieC View Post
    How do you integrate 1/(2v + 1)^2?
    Let u = 2v+1, then \dfrac{du}{dv} = 2 \Rightarrow dv = \dfrac{du}{2}. Thus \displaystyle \int\frac{1}{(2v+1)^2}\;{dv} = \frac{1}{2}\int\frac{1}{u^2}\;{du}.
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  4. #4
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    Use a u substitution.


    \int{\frac{1}{(2v + 1)^2}\,dv} = \frac{1}{2}\int{2(2v+1)^{-2}\,dv}.


    Now let u=2v + 1 so that \frac{du}{dv} = 2 and the integral becomes

    \frac{1}{2}\int{u^{-2}\,\frac{du}{dv}\,dv}

     = \frac{1}{2}\int{u^{-2}\,du}

     = \frac{1}{2}\left(\frac{u^{-1}}{-1}\right) + C

     = -\frac{1}{2u} + C

     = -\frac{1}{2(2v+1)} + C

     = -\frac{1}{4v+2} + C.
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