# Math Help - integral:completing the square

1. ## integral:completing the square

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

$\int\frac{1}{\sqrt{3-2x-x^2}}$

2. Hello,
Originally Posted by gammaman
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 : $4 \left(1-\left(\tfrac{x+1}{2}\right)^2\right)$