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Math Help - Integration

  1. #1
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    Integration

    Show that with limits 3 - 8, ∫(x+1)^1/2 / (x+5) dx = pi + 2 - 4 arctan 3/2.

    How can I do this?
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  2. #2
    Super Member wingless's Avatar
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    \int_{3}^{8}\frac{\sqrt{x+1}}{x+5}~dx

    Let t^2 = x+1. Then 2t~dt = dx.

    2\int_{2}^{3}\frac{t^2}{t^2+4}~dt

    To make it easier, I'll make one more substitution:

    2\int_{2}^{3}\frac{{\left(\frac{t}{2}\right)}^2}{{  \left(\frac{t}{2}\right)}^2+1}~dt

    k = \frac{t}{2} and it becomes,

    4\int_{1}^{3/2}\frac{k^2}{k^2+1}~dk

    Now use integration by parts to get the answer.
    u=k, dv = \frac{k}{k^2+1}~dk

    There surely must be a shorter way. This is only a brute force attempt to get the integral. I'll think on it again as soon as I have some time (if someone doesn't help you in that time ;p)
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  3. #3
    MHF Contributor
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    Rewrite just a little.

    \frac{\sqrt{x+1}}{(x+1)+4}

    \frac{Stuff}{Stuff^{2}+4}

    This should suggest a tantalizing substitution.
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