Refresh my memory, How would I go about completing the square so that I can solve this integral? $\displaystyle \int\frac{1}{\sqrt{3-2x-x^2}}$
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Hello, Originally Posted by gammaman Refresh my memory, How would I go about completing the square so that I can solve this integral? $\displaystyle \int\frac{1}{\sqrt{3-2x-x^2}}$ $\displaystyle 3-2x-x^2=-(x^2+2x-3)=-[x^2+2 \cdot {\color{red}1} \cdot x+{\color{red}1}^2-1-3]$ $\displaystyle =-[(x+1)^2-4]=4-(x+1)^2$ In order to solve the integral, write this as : $\displaystyle 4 \left(1-\left(\tfrac{x+1}{2}\right)^2\right)$
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