Quadratic integral

In mathematics, a quadratic integral of the form

$\int_{}^{}\ \frac{dx}{a+bx+cx^2}$

may be computed by completing the square in the denominator.

$\int_{}^{}\ \frac{dx}{a+bx+cx^2} = \frac{1}{c} \int_{}^{}\ \frac{dx}{\left( x+ \frac{b}{2c} \right)^2 + \left( \frac{a}{c} - \frac{b^2}{4c^2} \right)}$

Letting

$u \equiv x + \frac{b}{2c}$ ,

define

$-A^2 \equiv \frac{a}{c} - \frac{b^2}{4c^2} = \frac{1}{4c^2} \left( 4ac - b^2 \right) \equiv \frac{1}{4c^2} q$

where

$q \equiv 4ac-b^2$

is the negative of the discriminant. When q < 0, then

$A = \frac{1}{2c} \sqrt{-q}$

By use of partial fraction decomposition,

$\frac{1}{c} \int_{}^{}\ \frac{du}{(u + A)(u - A)} = \frac{1}{c} \int_{}^{}\ \left( \frac{A_1}{(u + A)} + \frac{A_2}{(u - A)} \right) du$

$= \frac{(A_1 + A_2)u + A(A_2 - A_1)}{u^2 - A^2}$

Then

$\frac{1}{c} \int_{}^{}\ \left( - \frac{1}{2A} \frac{1}{u+A} + \frac{1}{2A}\frac{1}{u-A} \right)du$

$= \frac{1}{2Ac} \ln \left( \frac{u - A}{u + A} \right)$

and finally

$= \frac{1}{ \sqrt{-q}} \ln \left( \frac{2cx + b - \sqrt{-q}}{2cx+b+ \sqrt{-q}} \right)$

Categories: Calculus