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Math Help - Don't know where to start with this integral

  1. #1
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    Don't know where to start with this integral

    I used to think I was pretty good at basic integration, then I met this

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    \int^8_3\frac{(x+1)^{\frac{1}{2}}}{x+5} \, dx = \pi + 2 + \tan^{-1}\frac{3}{2}

    Any advice on how to even start this one would be much appreciated. I think I'm missing a trick somewhere...
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  2. #2
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    Quote Originally Posted by Stonehambey View Post
    I used to think I was pretty good at basic integration, then I met this

    Show that

    \int^8_3\frac{(x+1)^{\frac{1}{2}}}{x+5} \, dx = \pi + 2 + \tan^{-1}\frac{3}{2}

    Any advice on how to even start this one would be much appreciated. I think I'm missing a trick somewhere...
    Try letting x + 1 = 4 \tan^2 \theta.
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  3. #3
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    It looks to me like it is coming from this formula

    \frac{1}{x^2+a^2}dx = \frac{1}{a}tan^{-1}\frac{x}{a}+c
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  4. #4
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    Let u=\sqrt{x+1}, so that x=u^2-1, dx=2u\,du, and the integral becomes:
    \int_3^8\frac{(x+1)^{\frac{1}{2}}}{x+5}\,dx=\int_2  ^3\frac{u}{u^2-1+5}\cdot2u\,du=2\int_2^3\frac{u^2}{u^2+4}\,du
    =2\int_2^3\frac{(u^2+4)-4}{u^2+4}\,du
    =2\left[\int_2^3\,du-4\int_2^3\frac{1}{u^2+4}\,du\right]
    =2-8\int_2^3\frac{1}{u^2+4}\,du
    Use the inverse tangent formula gammaman gave you for this last integral, and you should get the result, particularly if you recall that \tan^{-1}(1)=\frac{\pi}{4}.

    --Kevin C.
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