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

${\displaystyle \int {\frac {dx}{a+bx+cx^{2}}}.}$

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

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

## Positive-discriminant case

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

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

and

${\displaystyle -A^{2}={\frac {a}{c}}-{\frac {b^{2}}{4c^{2}}}={\frac {1}{4c^{2}}}\left(4ac-b^{2}\right).}$

The quadratic integral can now be written as

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

allows us to evaluate the integral:

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

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

${\displaystyle \int {\frac {dx}{a+bx+cx^{2}}}={\frac {1}{\sqrt {q}}}\ln \left({\frac {2cx+b-{\sqrt {q}}}{2cx+b+{\sqrt {q}}}}\right)+{\text{constant, where }}q=b^{2}-4ac.}$

## 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

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

is positive. Then the integral becomes

{\displaystyle {\begin{aligned}&{}\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{aligned}}}