Quadratic integral

In mathematics, a quadratic integral is an integral of the form

∫ d x a + b x + c x 2 . {\displaystyle \int {\frac {dx}{a+bx+cx^{2}}}.}

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

∫ d x a + b x + c x 2 = 1 c ∫ d x ( x + b 2 c ) 2 + ( a c − b 2 4 c 2 ) . {\displaystyle \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)}}.}

Positive-discriminant case

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

u = x + b 2 c , {\displaystyle u=x+{\frac {b}{2c}},}
and
− A 2 = a c − b 2 4 c 2 = 1 4 c 2 ( 4 a c − b 2 ) . {\displaystyle -A^{2}={\frac {a}{c}}-{\frac {b^{2}}{4c^{2}}}={\frac {1}{4c^{2}}}(4ac-b^{2}).}

The quadratic integral can now be written as

∫ d x a + b x + c x 2 = 1 c ∫ d u u 2 − A 2 = 1 c ∫ d u ( u + A ) ( u − A ) . {\displaystyle \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)}}.}

The partial fraction decomposition

1 ( u + A ) ( u − A ) = 1 2 A ( 1 u − A − 1 u + A ) {\displaystyle {\frac {1}{(u+A)(u-A)}}={\frac {1}{2A}}\!\left({\frac {1}{u-A}}-{\frac {1}{u+A}}\right)}
allows us to evaluate the integral:
1 c ∫ d u ( u + A ) ( u − A ) = 1 2 A c ln ⁡ ( u − A u + A ) + constant . {\displaystyle {\frac {1}{c}}\int {\frac {du}{(u+A)(u-A)}}={\frac {1}{2Ac}}\ln \left({\frac {u-A}{u+A}}\right)+{\text{constant}}.}

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

∫ d x a + b x + c x 2 = 1 q ln ⁡ ( 2 c x + b − q 2 c x + b + q ) + constant . {\displaystyle \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}}.}

Negative-discriminant case

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

∫ d x a + b x + c x 2 = 1 c ∫ d x ( x + b 2 c ) 2 + ( a c − b 2 4 c 2 ) . {\displaystyle \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)}}.}
is positive. Then the integral becomes
1 c ∫ d u u 2 + A 2 = 1 c A ∫ d u / A ( u / A ) 2 + 1 = 1 c A ∫ d w w 2 + 1 = 1 c A arctan ⁡ ( w ) + c o n s t a n t = 1 c A arctan ⁡ ( u A ) + constant = 1 c a c − b 2 4 c 2 arctan ⁡ ( x + b 2 c a c − b 2 4 c 2 ) + constant = 2 4 a c − b 2 arctan ⁡ ( 2 c x + b 4 a c − b 2 ) + constant . {\displaystyle {\begin{aligned}{\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{aligned}}}

References

  • Weisstein, Eric W. "Quadratic Integral." From MathWorld--A Wolfram Web Resource, wherein the following is referenced:
  • Gradshteyn, Izrail Solomonovich; Ryzhik, Iosif Moiseevich; Geronimus, Yuri Veniaminovich; Tseytlin, Michail Yulyevich; Jeffrey, Alan (2015) [October 2014]. Zwillinger, Daniel; Moll, Victor Hugo (eds.). Table of Integrals, Series, and Products. Translated by Scripta Technica, Inc. (8 ed.). Academic Press, Inc. ISBN 978-0-12-384933-5. LCCN 2014010276.
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