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[ | In [[mathematics]], a '''quadratic integral''' is an [[integral]] of the form | ||
:<math>\int \frac{dx}{a+bx+cx^2}. </math> | |||
It can be evaluated by [[completing the square]] in the [[denominator]]. | |||
:<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> | |||
==Positive-discriminant case== | |||
Assume that the [[discriminant]] ''q'' = ''b''<sup>2</sup> − 4''ac'' is positive. In that case, define ''u'' and ''A'' by | |||
:<math>u = x + \frac{b}{2c} </math>, | |||
and | |||
:<math> -A^2 = \frac{a}{c} - \frac{b^2}{4c^2} = \frac{1}{4c^2} \left( 4ac - b^2 \right). </math> | |||
The quadratic integral can now be written as | |||
:<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> | |||
The [[partial fraction decomposition]] | |||
:<math> \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> \frac1c \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> \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> | |||
==Negative-discriminant case== | |||
:''This (hastily written) section may need attention.'' | |||
In case the [[discriminant]] ''q'' = ''b''<sup>2</sup> − 4''ac'' is negative, the second term in the denominator in | |||
:<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> | |||
is positive. Then the integral becomes | |||
:<math> | |||
\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} | |||
</math> | |||
==References== | |||
*Weisstein, Eric W. "[http://mathworld.wolfram.com/QuadraticIntegral.html 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. | |||
[[Category:Integral calculus]] |
Revision as of 08:53, 31 January 2014
In mathematics, a quadratic integral is an integral of the form
It can be evaluated by completing the square in the denominator.
Positive-discriminant case
Assume that the discriminant q = b2 − 4ac is positive. In that case, define u and A by
and
The quadratic integral can now be written as
The partial fraction decomposition
allows us to evaluate the integral:
The final result for the original integral, under the assumption that q > 0, is
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
is positive. Then the integral becomes
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.