Quadratic integral
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In mathematics, a quadratic integral of the form
- <math>\int_{}^{}\ \frac{dx}{a+bx+cx^2} <math>
may be computed by completing the square in the denominator.
- <math>\int_{}^{}\ \frac{dx}{a+bx+cx^2} = \frac{1}{c} \int_{}^{}\ \frac{dx}{\left( x+ \frac{b}{2c} \right)^2 + \left( \frac{a}{c} - \frac{b^2}{4c^2} \right)}<math>
Letting
- <math>u \equiv x + \frac{b}{2c} <math>,
define
- <math> -A^2 \equiv \frac{a}{c} - \frac{b^2}{4c^2} = \frac{1}{4c^2} \left( 4ac - b^2 \right) \equiv \frac{1}{4c^2} q <math>
where
- <math> q \equiv 4ac-b^2<math>
is the negative of the discriminant. When q < 0, then
- <math> A = \frac{1}{2c} \sqrt{-q}<math>
By use of partial fraction decomposition,
- <math> \frac{1}{c} \int_{}^{}\ \frac{du}{(u + A)(u - A)} = \frac{1}{c} \int_{}^{}\ \left( \frac{A_1}{(u + A)} + \frac{A_2}{(u - A)} \right) du <math>
- <math> = \frac{(A_1 + A_2)u + A(A_2 - A_1)}{u^2 - A^2}<math>
Then
- <math> \frac{1}{c} \int_{}^{}\ \left( - \frac{1}{2A} \frac{1}{u+A} + \frac{1}{2A}\frac{1}{u-A} \right)du<math>
- <math> = \frac{1}{2Ac} \ln \left( \frac{u - A}{u + A} \right)<math>
and finally
- <math> = \frac{1}{ \sqrt{-q}} \ln \left( \frac{2cx + b - \sqrt{-q}}{2cx+b+ \sqrt{-q}} \right) <math>