In mathematics , a quadratic integral is an integral of the form
∫
d
x
a
+
b
x
+
c
x
2
.
{\displaystyle \int {\frac {dx}{a+bx+cx^{2}}}.}
It can be evaluated by completing the square in the denominator .
∫
d
x
a
+
b
x
+
c
x
2
=
1
c
∫
d
x
(
x
+
b
2
c
)
2
+
(
a
c
−
b
2
4
c
2
)
.
{\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 = b − 4ac is positive. In that case, define u and A by
u
=
x
+
b
2
c
,
{\displaystyle u=x+{\frac {b}{2c}},}
−
A
2
=
a
c
−
b
2
4
c
2
=
1
4
c
2
(
4
a
c
−
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
∫
d
x
a
+
b
x
+
c
x
2
=
1
c
∫
d
u
u
2
−
A
2
=
1
c
∫
d
u
(
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
1
(
u
+
A
)
(
u
−
A
)
=
1
2
A
(
1
u
−
A
−
1
u
+
A
)
{\displaystyle {\frac {1}{(u+A)(u-A)}}={\frac {1}{2A}}\!\left({\frac {1}{u-A}}-{\frac {1}{u+A}}\right)}
1
c
∫
d
u
(
u
+
A
)
(
u
−
A
)
=
1
2
A
c
ln
(
u
−
A
u
+
A
)
+
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
∫
d
x
a
+
b
x
+
c
x
2
=
1
q
ln
(
2
c
x
+
b
−
q
2
c
x
+
b
+
q
)
+
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}}.}
Negative-discriminant case
In case the discriminant q = b − 4ac is negative, the second term in the denominator in
∫
d
x
a
+
b
x
+
c
x
2
=
1
c
∫
d
x
(
x
+
b
2
c
)
2
+
(
a
c
−
b
2
4
c
2
)
.
{\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)}}.}
1
c
∫
d
u
u
2
+
A
2
=
1
c
A
∫
d
u
/
A
(
u
/
A
)
2
+
1
=
1
c
A
∫
d
w
w
2
+
1
=
1
c
A
arctan
(
w
)
+
c
o
n
s
t
a
n
t
=
1
c
A
arctan
(
u
A
)
+
constant
=
1
c
a
c
−
b
2
4
c
2
arctan
(
x
+
b
2
c
a
c
−
b
2
4
c
2
)
+
constant
=
2
4
a
c
−
b
2
arctan
(
2
c
x
+
b
4
a
c
−
b
2
)
+
constant
.
{\displaystyle {\begin{aligned}{\frac {1}{c}}\int {\frac {du}{u^{2}+A^{2}}}&={\frac {1}{cA}}\int {\frac {du/A}{(u/A)^{2}+1}}\\&={\frac {1}{cA}}\int {\frac {dw}{w^{2}+1}}\\&={\frac {1}{cA}}\arctan(w)+\mathrm {constant} \\&={\frac {1}{cA}}\arctan \left({\frac {u}{A}}\right)+{\text{constant}}\\&={\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}}\\&={\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, Izrail Solomonovich ; Ryzhik, Iosif Moiseevich ; Geronimus, Yuri Veniaminovich ; Tseytlin, Michail Yulyevich ; Jeffrey, Alan (2015) . Zwillinger, Daniel; Moll, Victor Hugo (eds.). Table of Integrals, Series, and Products Academic Press, Inc. ISBN 978-0-12-384933-5 LCCN 2014010276 .Category :
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and
allows us to evaluate the integral:
