Inmathematics, aquadratic integral is anintegral of the form$${\displaystyle \int \frac{dx}{a+bx+c{x}^2}.}$$

It can be evaluated bycompleting the square in thedenominator.

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

Assume that thediscriminantq =b² − 4ac is positive. In that case, defineu andA by$${\displaystyle u=x+\frac{b}{2c},}$$and$${\displaystyle -A^2=\frac{a}{c}-\frac{b^2}{4c^2}=\frac{1}{4c^2}(4ac-b^2).}$$

The quadratic integral can now be written as$${\displaystyle \int \frac{dx}{a+bx+c{x}^2}=\frac{1}{c}\int \frac{du}{u^2-A^2}=\frac{1}{c}\int \frac{du}{(u+A)(u-A)}.}$$

Thepartial fraction decomposition$${\displaystyle \frac{1}{(u+A)(u-A)}=\frac{1}{2A}\left(\frac{1}{u-A}-\frac{1}{u+A}\right)}$$allows us to evaluate the integral:$${\displaystyle \frac{1}{c}\int \frac{du}{(u+A)(u-A)}=\frac{1}{2Ac}\ln \left(\frac{u-A}{u+A}\right)+\text{constant}.}$$

The final result for the original integral, under the assumption thatq > 0, is$${\displaystyle \int \frac{dx}{a+bx+c{x}^2}=\frac{1}{\sqrt{q}}\ln \left(\frac{2cx+b-\sqrt{q}}{2cx+b+\sqrt{q}}\right)+\text{constant}.}$$

In case thediscriminantq =b² − 4ac is negative, the second term in the denominator in$${\displaystyle \int \frac{dx}{a+bx+c{x}^2}=\frac{1}{c}\int \frac{dx}{{\left(x+\frac{b}{2c}\right)}^2+\left(\frac{a}{c}-\frac{b^2}{4c^2}\right)}.}$$is positive. Then the integral becomes

• Weisstein, Eric W. "Quadratic Integral." FromMathWorld--A Wolfram Web Resource, wherein the following is referenced:
• Gradshteyn, Izrail Solomonovich;Ryzhik, Iosif Moiseevich;Geronimus, Yuri Veniaminovich;Tseytlin, Michail Yulyevich; Jeffrey, Alan (2015) [October 2014]. Zwillinger, Daniel;Moll, Victor Hugo (eds.).Table of Integrals, Series, and Products. Translated by Scripta Technica, Inc. (8 ed.).Academic Press, Inc.ISBN 978-0-12-384933-5.LCCN 2014010276.

• Integral calculus