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Math Help - integral:completing the square

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    integral:completing the square

    Refresh my memory,
    How would I go about completing the square so that I can solve this integral?

    \int\frac{1}{\sqrt{3-2x-x^2}}
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    Hello,
    Quote Originally Posted by gammaman View Post
    Refresh my memory,
    How would I go about completing the square so that I can solve this integral?

    \int\frac{1}{\sqrt{3-2x-x^2}}
    3-2x-x^2=-(x^2+2x-3)=-[x^2+2 \cdot {\color{red}1} \cdot x+{\color{red}1}^2-1-3]
    =-[(x+1)^2-4]=4-(x+1)^2

    In order to solve the integral, write this as : 4 \left(1-\left(\tfrac{x+1}{2}\right)^2\right)
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