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Math Help - problematic integral #2

  1. #1
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    problematic integral #2

    Hi.
    Evaluate:

    \int <br />
{\sqrt{x^2+x+3} dx}
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  2. #2
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    Quote Originally Posted by Shock View Post
    Hi.
    Evaluate:

    \int <br />
{\sqrt{x^2+x+3} dx}
    See here: integrate Sqrt&#x5b;x&#x5e;2 &#x2b; x &#x2b; 3&#x5d; - Wolfram|Alpha

    (Click on Show steps).
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  3. #3
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    Hello,

    By completing the square, we have x^2+x+3=\left(x+\tfrac 12\right)^2+\tfrac{11}{4}=\tfrac{11}{4}\left(\tfra  c{4}{11}\cdot\left(x+\tfrac 12\right)^2+1\right)

    Using the usual substitutions, let \sinh(u)=\sqrt{\tfrac{4}{11}}\cdot\left(x+\tfrac 12\right)

    (thus \cosh(u) ~du=\sqrt{\tfrac{4}{11}} ~dx)

    And you'll have \int \sqrt{\tfrac{11}{4}}\cdot\sqrt{\tfrac{11}{4}} \cdot \sqrt{\sinh^2(u)+1}\cdot\cosh(u) ~du=\tfrac{11}{4}\int \cosh^2(u) ~du

    But 2\cosh^2(u)=1+\cosh(2u)

    So \tfrac{11}{4}\int \cosh^2(u) ~du=\tfrac{11}{8}\left(x+\tfrac 12 \cdot \sinh(2u)\right)+C

    Now, back substitute
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