Editing Quadratic integral (section)

==Positive-discriminant case==

Assume that the [[discriminant]] ''q'' = ''b''<sup>2</sup>&nbsp;−&nbsp;4''ac'' is positive. In that case, define ''u'' and ''A'' by
<math display="block">u = x + \frac{b}{2c},</math>
and
<math display="block">-A^2 = \frac{a}{c} - \frac{b^2}{4c^2} = \frac{1}{4c^2}(4ac - b^2).</math>

The quadratic integral can now be written as
<math display="block">\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)}.</math>

The [[partial fraction decomposition]]
<math display="block">\frac{1}{(u+A)(u-A)} = \frac{1}{2A}\!\left( \frac{1}{u-A} - \frac{1}{u+A} \right) </math>
allows us to evaluate the integral:
<math display="block">\frac{1}{c} \int \frac{du}{(u+A)(u-A)} = \frac{1}{2Ac} \ln \left( \frac{u - A}{u + A} \right) + \text{constant}.</math>

The final result for the original integral, under the assumption that ''q'' > 0, is
<math display="block">\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}.</math>