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

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

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

The quadratic integral can now be written as

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

The partial fraction decomposition

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

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

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

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

Negative-discriminant case

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

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

$${\displaystyle {\begin{aligned}{\frac {1}{c))\int {\frac {du}{u^{2}+A^{2))}&={\frac {1}{cA))\int {\frac {du/A}{(u/A)^{2}+1))\\[9pt]&={\frac {1}{cA))\int {\frac {dw}{w^{2}+1))\\[9pt]&={\frac {1}{cA))\arctan(w)+\mathrm {constant} \\[9pt]&={\frac {1}{cA))\arctan \left({\frac {u}{A))\right)+{\text{constant))\\[9pt]&={\frac {1}{c{\sqrt ((\frac {a}{c))-{\frac {b^{2)){4c^{2))))))}\arctan \left({\frac {x+{\frac {b}{2c))}{\sqrt ((\frac {a}{c))-{\frac {b^{2)){4c^{2))))))\right)+{\text{constant))\\[9pt]&={\frac {2}{\sqrt {4ac-b^{2}\,))}\arctan \left({\frac {2cx+b}{\sqrt {4ac-b^{2))))\right)+{\text{constant)).\end{aligned))}$$