Faraday effect

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

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

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

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

Positive-discriminant case

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

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

and

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

The quadratic integral can now be written as

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

The partial fraction decomposition

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

allows us to evaluate the integral:

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

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

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

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

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

is positive. Then the integral becomes

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

References

• Weisstein, Eric W. "Quadratic Integral." From MathWorld--A Wolfram Web Resource, wherein the following is referenced:
• Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 6th ed. San Diego, CA: Academic Press, 2000.