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Math Help - Simple antiderivatives

  1. #1
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    Simple antiderivatives

    Okay, I'm doing Integral tests for series and I think my brain is just off cause I can't find these simple antiderivatives. My first thought was make it an negative exponent, but yeah, that doesn't really help anything. Can someone just explain one to me cause I can't even think how to start these?

    anti 1/(2n - 1)
    anti 1/(n^2 + 4)
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by DemonGal711 View Post
    Okay, I'm doing Integral tests for series and I think my brain is just off cause I can't find these simple antiderivatives. My first thought was make it an negative exponent, but yeah, that doesn't really help anything. Can someone just explain one to me cause I can't even think how to start these?

    anti 1/(2n - 1)
    anti 1/(n^2 + 4)
    The first one:

    We will use u-substitution.

    \int \frac{1}{2n-1} \ dn

    --
    u = 2n - 1
    du = 2n \rightarrow \frac{1}{2} \ du = dn
    --

    \frac{1}{2} \int \frac{1}{u} \ du

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

    = \frac{1}{2} \ln{2n - 1} + C


    _____________


    The second one:

    \int \frac{1}{4 + n^2} \ dn

    You'll have to study the forms of these integrals so you can easily spot this one is going to be arctan.

    For an integral of this form:
    \int \frac{1}{a^2 + u^2} \ du = \frac{1}{a} tan^{-1} \left( \frac{u}{a} \right) + C

    So in your case we have:

    \int \frac{1}{4 + n^2} \ dn = \frac{1}{2} tan^{-1} \left( \frac{n}{2} \right) + C
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