Quadratic integral

In mathematics, a quadratic integral is an integral of the form $${\displaystyle \int {\frac {dx}{a+bx+cx^{2}}}.}$${\displaystyle \int {\frac {dx}{a+bx+cx^{2}}}.}

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

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

Assume that the discriminant q = b² − 4ac is positive. In that case, define u and A by $${\displaystyle u=x+{\frac {b}{2c}},}$${\displaystyle u=x+{\frac {b}{2c}},} and $${\displaystyle -A^{2}={\frac {a}{c}}-{\frac {b^{2}}{4c^{2}}}={\frac {1}{4c^{2}}}(4ac-b^{2}).}$${\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 $${\displaystyle \int {\frac {dx}{a+bx+cx^{2}}}={\frac {1}{c}}\int {\frac {du}{u^{2}-A^{2}}}={\frac {1}{c}}\int {\frac {du}{(u+A)(u-A)}}.}$${\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 $${\displaystyle {\frac {1}{(u+A)(u-A)}}={\frac {1}{2A}}\!\left({\frac {1}{u-A}}-{\frac {1}{u+A}}\right)}$${\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: $${\displaystyle {\frac {1}{c}}\int {\frac {du}{(u+A)(u-A)}}={\frac {1}{2Ac}}\ln \left({\frac {u-A}{u+A}}\right)+{\text{constant}}.}$${\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 $${\displaystyle \int {\frac {dx}{a+bx+cx^{2}}}={\frac {1}{\sqrt {q}}}\ln \left({\frac {2cx+b-{\sqrt {q}}}{2cx+b+{\sqrt {q}}}}\right)+{\text{constant}}.}$${\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}}.}

In case the discriminant q = b² − 4ac is negative, the second term in the denominator in $${\displaystyle \int {\frac {dx}{a+bx+cx^{2}}}={\frac {1}{c}}\int {\frac {dx}{\left(x+{\frac {b}{2c}}\right)^{\!2}+\left({\frac {a}{c}}-{\frac {b^{2}}{4c^{2}}}\right)}}.}$${\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 $${\displaystyle {\begin{aligned}{\frac {1}{c}}\int {\frac {du}{u^{2}+A^{2}}}&={\frac {1}{cA}}\int {\frac {du/A}{(u/A)^{2}+1}}\\[9pt]&={\frac {1}{cA}}\int {\frac {dw}{w^{2}+1}}\\[9pt]&={\frac {1}{cA}}\arctan(w)+\mathrm {constant} \\[9pt]&={\frac {1}{cA}}\arctan \left({\frac {u}{A}}\right)+{\text{constant}}\\[9pt]&={\frac {1}{c{\sqrt {{\frac {a}{c}}-{\frac {b^{2}}{4c^{2}}}}}}}\arctan \left({\frac {x+{\frac {b}{2c}}}{\sqrt {{\frac {a}{c}}-{\frac {b^{2}}{4c^{2}}}}}}\right)+{\text{constant}}\\[9pt]&={\frac {2}{\sqrt {4ac-b^{2},円}}}\arctan \left({\frac {2cx+b}{\sqrt {4ac-b^{2}}}}\right)+{\text{constant}}.\end{aligned}}}$${\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},円}}}\arctan \left({\frac {2cx+b}{\sqrt {4ac-b^{2}}}}\right)+{\text{constant}}.\end{aligned}}}

• 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.

• Integral calculus