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| In [[mathematics]], a '''quadratic integral''' is an [[integral]] of the form
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| :<math>\int \frac{dx}{a+bx+cx^2}. </math> | |
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| It can be evaluated by [[completing the square]] in the [[denominator]].
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| :<math>\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>
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| ==Positive-discriminant case== | |
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| Assume that the [[discriminant]] ''q'' = ''b''<sup>2</sup> − 4''ac'' is positive. In that case, define ''u'' and ''A'' by
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| :<math>u = x + \frac{b}{2c} </math>,
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| :<math> -A^2 = \frac{a}{c} - \frac{b^2}{4c^2} = \frac{1}{4c^2} \left( 4ac - b^2 \right). </math>
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| The quadratic integral can now be written as
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| :<math> \int \frac{dx}{a+bx+cx^2} = \frac1c \int \frac{du}{u^2-A^2} = \frac1c \int \frac{du}{(u+A)(u-A)}. </math>
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| The [[partial fraction decomposition]]
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| :<math> \frac{1}{(u+A)(u-A)} = \frac{1}{2A} \left( \frac{1}{u-A} - \frac{1}{u+A} \right) </math>
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| allows us to evaluate the integral:
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| :<math> \frac1c \int \frac{du}{(u+A)(u-A)} = \frac{1}{2Ac} \ln \left( \frac{u - A}{u + A} \right) + \text{constant}. </math> | |
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| The final result for the original integral, under the assumption that ''q'' > 0, is
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| :<math> \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. </math>
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| ==Negative-discriminant case==
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| :''This (hastily written) section may need attention.''
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| In case the [[discriminant]] ''q'' = ''b''<sup>2</sup> − 4''ac'' is negative, the second term in the denominator in
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| :<math>\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>
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| is positive. Then the integral becomes
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| :<math>
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| \begin{align}
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| & {} \qquad \frac{1}{c} \int \frac{ du} {u^2 + A^2} \\[9pt]
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| & = \frac{1}{cA} \int \frac{du/A}{(u/A)^2 + 1 } \\[9pt]
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| & = \frac{1}{cA} \int \frac{dw}{w^2 + 1} \\[9pt]
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| & = \frac{1}{cA} \arctan(w) + \mathrm{constant} \\[9pt]
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| & = \frac{1}{cA} \arctan\left(\frac{u}{A}\right) + \text{constant} \\[9pt]
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| & = \frac{1}{c\sqrt{\frac{a}{c} - \frac{b^2}{4c^2}}} \arctan
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| \left(\frac{x + \frac{b}{2c}}{\sqrt{\frac{a}{c} - \frac{b^2}{4c^2}}}\right) + \text{constant} \\[9pt]
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| & = \frac{2}{\sqrt{4ac - b^2\, }}
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| \arctan\left(\frac{2cx + b}{\sqrt{4ac - b^2}}\right) + \text{constant}.
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| \end{align}
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| </math>
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| ==References==
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| *Weisstein, Eric W. "[http://mathworld.wolfram.com/QuadraticIntegral.html Quadratic Integral]." From ''MathWorld''--A Wolfram Web Resource, wherein the following is referenced:
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| *Gradshteyn, I. S. and Ryzhik, I. M. ''Tables of Integrals, Series, and Products,'' 6th ed. San Diego, CA: Academic Press, 2000.
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| [[Category:Integral calculus]]
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