Quadratic integral

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

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

and

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

The quadratic integral can now be written as

$$\backslash int\; \backslash frac\{dx\}\{a+bx+cx^2\}\; =\; \backslash frac\{1\}\{c\}\; \backslash int\; \backslash frac\{du\}\{u^2-A^2\}\; =\; \backslash frac\{1\}\{c\}\; \backslash int\; \backslash frac\{du\}\{(u+A)(u-A)\}.$$

The partial fraction decomposition

$$\backslash frac\{1\}\{(u+A)(u-A)\}\; =\; \backslash frac\{1\}\{2A\}\backslash !\backslash left(\; \backslash frac\{1\}\{u-A\}\; -\; \backslash frac\{1\}\{u+A\}\; \backslash right)$$

allows us to evaluate the integral:

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

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

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

In case the discriminant q = b² − 4ac is negative, the second term in the denominator in

$$\backslash int\; \backslash frac\{dx\}\{a+bx+cx^2\}\; =\; \backslash frac\{1\}\{c\}\; \backslash int\; \backslash frac\{dx\}\{\backslash left(\; x+\; \backslash frac\{b\}\{2c\}\; \backslash right)^\{\backslash !2\}\; +\; \backslash left(\; \backslash frac\{a\}\{c\}\; -\; \backslash frac\{b^2\}\{4c^2\}\; \backslash right)\}.$$

is positive. Then the integral becomes

$$\backslash begin\{align\}$$

\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}

• 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 |title-link=Gradshteyn and Ryzhik}}

{{Calculus topics}}

Category:Integral calculus