Overview
In mathematics, a quadratic integral is an integral of the form \int \frac dx a+bx+cx^2 . It can be evaluated 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) .
Positive-discriminant case
Assume that the discriminant q = b 2 − 4ac is positive. In that case, define u and A by u = x + \frac b 2c , and -A^2 = \frac a c - \frac b^2 4c^2 = \frac 1 4c^2 (4ac - b^2). The quadratic integral can now be written as \int \frac dx a+bx+cx^2 = \frac 1 c \int \frac du u^2-A^2 = \frac 1 c \int \frac du (u+A)(u-A) . The partial fraction decomposition \frac 1 (u+A)(u-A) = \frac 1 2A \!\left( \frac 1 u-A - \frac 1 u+A \right) allows us to evaluate the integral: \frac 1 c \int \frac du (u+A)(u-A) = \frac 1 2Ac \ln \left( \frac u - A u + A \right) + \text constant . The final result for the original integral, under the assumption that q > 0, is \int \frac dx a+bx+cx^2 = \frac 1 \sqrt q \ln \left( \frac 2cx + b - \sqrt q 2cx+b+ \sqrt q \right) + \text constant .
Negative-discriminant case
In case the discriminant q = b 2 − 4ac is negative, the second term in the denominator in \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) . is positive. Then the integral becomes \begin align \frac 1 c \int \frac du u^2 + A^2 & = \frac 1 cA \int \frac du/A (u/A)^2 + 1 \\ 9pt & = \frac 1 cA \int \frac dw w^2 + 1 \\ 9pt & = \frac 1 cA \arctan(w) + \mathrm constant \\ 9pt & = \frac 1 cA \arctan\left(\frac u A \right) + \text constant \\ 9pt & = \frac 1 c\sqrt \frac a c - \frac b^2 4c^2 \arctan \left(\frac x + \frac b 2c \sqrt \frac a c - \frac b^2 4c^2 \right) + \text constant \\ 9pt & = \frac 2 \sqrt 4ac - b^2\, \arctan\left(\frac 2cx + b \sqrt 4ac - b^2 \right) + \text constant . \end align
References
*Weisstein, Eric W. "Quadratic Integral." From MathWorld--A Wolfram Web Resource, wherein the following is referenced: *