Editing 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'' = ''b''<sup>2</sup>&nbsp;−&nbsp;4''ac'' 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'' = ''b''<sup>2</sup>&nbsp;−&nbsp;4''ac'' 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. "[http://mathworld.wolfram.com/QuadraticIntegral.html Quadratic Integral]." From ''MathWorld''--A Wolfram Web Resource, wherein the following is referenced:
*{{cite book |author-first1=Izrail Solomonovich |author-last1=Gradshteyn |author-link1=Izrail Solomonovich Gradshteyn |author-first2=Iosif Moiseevich |author-last2=Ryzhik |author-link2=Iosif Moiseevich Ryzhik |author-first3=Yuri Veniaminovich |author-last3=Geronimus |author-link3=Yuri Veniaminovich Geronimus |author-first4=Michail Yulyevich |author-last4=Tseytlin |author-link4=Michail Yulyevich Tseytlin |author-first5=Alan |author-last5=Jeffrey |editor-first1=Daniel |editor-last1=Zwillinger |editor-first2=Victor Hugo |editor-last2=Moll |editor-link2=Victor Hugo Moll |translator=Scripta Technica, Inc. |title=Table of Integrals, Series, and Products |publisher=[[Academic Press, Inc.]] |date=2015 |orig-year=October 2014 |edition=8 |language=English |isbn=978-0-12-384933-5 |lccn=2014010276 <!-- |url=https://books.google.com/books?id=NjnLAwAAQBAJ |access-date=2016-02-21-->|title-link=Gradshteyn and Ryzhik}}

{{Calculus topics}}

[[Category:Integral calculus]]