Quadratic integral

In mathematics, a quadratic integral is an integral of the form <math display="block">\int \frac{dx}{a+bx+cx^2}.</math>

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

<math display="block">\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 display="block">u = x + \frac{b}{2c},</math> and <math display="block">-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 display="block">\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 display="block">\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 display="block">\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 display="block">\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 display="block">\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 display="block">\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:
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