Quadratic integral

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 = b² − 4ac is positive. In that case, define u and A by

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

and

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

The quadratic integral can now be written as

$\int \frac{dx}{a+bx+cx^2} = \frac1c \int \frac{du}{u^2-A^2} = \frac1c \int \frac{du}{(u+A)(u-A)}.$

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:

$\frac1c \int \frac{du}{(u+A)(u-A)} = \frac{1}{2Ac} \ln \left( \frac{u - A}{u + A} \right) + \text{constant}.$

The final result for the original integral, under the assumption that q > 0, is

$\int \frac{dx}{a+bx+cx^2} = \frac{1}{ \sqrt{q}} \ln \left( \frac{2cx + b - \sqrt{q}}{2cx+b+ \sqrt{q}} \right) + \text{constant, where } q = b^2 - 4ac.$

Negative-discriminant case

This (hastily written) section may need attention.

In case the discriminant q = b² − 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

$\begin{align} & {} \qquad \frac{1}{c} \int \frac{ du} {u^2 + A^2} \\[9pt] & = \frac{1}{cA} \int \frac{du/A}{(u/A)^2 + 1 } \\[9pt] & = \frac{1}{cA} \int \frac{dw}{w^2 + 1} \\[9pt] & = \frac{1}{cA} \arctan(w) + \mathrm{constant} \\[9pt] & = \frac{1}{cA} \arctan\left(\frac{u}{A}\right) + \text{constant} \\[9pt] & = \frac{1}{c\sqrt{\frac{a}{c} - \frac{b^2}{4c^2}}} \arctan \left(\frac{x + \frac{b}{2c}}{\sqrt{\frac{a}{c} - \frac{b^2}{4c^2}}}\right) + \text{constant} \\[9pt] & = \frac{2}{\sqrt{4ac - b^2\, }} \arctan\left(\frac{2cx + b}{\sqrt{4ac - b^2}}\right) + \text{constant}. \end{align}$

References

• Weisstein, Eric W. "Quadratic Integral." From MathWorld--A Wolfram Web Resource, wherein the following is referenced:
• Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 6th ed. San Diego, CA: Academic Press, 2000.