Quadratic integral

In mathematics, a quadratic integral is an integral of the form [math]\displaystyle{ \int \frac{dx}{a+bx+cx^2}. }[/math]

It can be evaluated by completing the square in the denominator.

[math]\displaystyle{ \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]

Positive-discriminant case

Assume that the discriminant q = b² − 4ac is positive. In that case, define u and A by [math]\displaystyle{ u = x + \frac{b}{2c}, }[/math] and [math]\displaystyle{ -A^2 = \frac{a}{c} - \frac{b^2}{4c^2} = \frac{1}{4c^2}(4ac - b^2). }[/math]

The quadratic integral can now be written as [math]\displaystyle{ \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)}. }[/math]

The partial fraction decomposition [math]\displaystyle{ \frac{1}{(u+A)(u-A)} = \frac{1}{2A}\!\left( \frac{1}{u-A} - \frac{1}{u+A} \right) }[/math] allows us to evaluate the integral: [math]\displaystyle{ \frac{1}{c} \int \frac{du}{(u+A)(u-A)} = \frac{1}{2Ac} \ln \left( \frac{u - A}{u + A} \right) + \text{constant}. }[/math]

The final result for the original integral, under the assumption that q > 0, is [math]\displaystyle{ \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}. }[/math]

Negative-discriminant case

In case the discriminant q = b² − 4ac is negative, the second term in the denominator in [math]\displaystyle{ \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] is positive. Then the integral becomes [math]\displaystyle{ \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} }[/math]

References

• Weisstein, Eric W. "Quadratic Integral." From MathWorld--A Wolfram Web Resource, wherein the following is referenced:
• (in English) Table of Integrals, Series, and Products (8 ed.). Academic Press, Inc.. 2015. ISBN 978-0-12-384933-5.