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Math Help - hard integral

  1. #1
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    hard integral

    solve the following integral:

    \int \frac{4x + 4}{(x^2+1)^2} dx
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  2. #2
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    Quote Originally Posted by qzno View Post
    solve the following integral:

    \int \frac{4x + 4}{(x^2+1)^2} dx
    (x^2 + 1) = t

    2x dx = dt


    Now 4 x + 4 = 2(2x) + 4

    \int \frac{4x + 4}{(x^2+1)^2} dx

    \int (\frac{2(2x) }{(x^2+1)^2} +\frac{4}{(x^2 + 1)^2})dx

    \int \frac{2dt}{(t)^2} +\frac{4dx}{(x^2 + 1)^2}
    ---------------------------------------------------------------------------
    First one is easy

    Second one

    \frac{4dx}{(x^2 + 1)^2}  <br />

    Put~ x =tan(\theta)

    dx =sec^2 (\theta) d\theta

    Integration becomes

    \int \frac{4sec^2(\theta)d\theta}{(sec^2(\theta))^2}  <br />

    \int 4cos^2(\theta)d\theta<br />

    Total integration is

    \int \frac{2dt}{(t)^2} +\int 4cos^2(\theta)d\theta
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