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Quadratic Integral

To compute an integral of the form

\int {dx\over a+bx+cx^2},
\end{displaymath} (1)

Complete the Square in the Denominator to obtain
\int {dx\over a+bx+cx^2} = {1\over c}\int {dx\over \left({x+...
...ver 2c}}\right)^2 +\left({{a\over c}-{b^2\over 4c^2}}\right)}.
\end{displaymath} (2)

Let $u\equiv x+b/2c$. Then define
-A^2\equiv {a\over c}-{b^2\over 4c^2} = {1\over 4c^2}(4ac-b^2)\equiv{1\over 4c^2} q,
\end{displaymath} (3)

q\equiv 4ac-b^2
\end{displaymath} (4)

is the Negative of the Discriminant. If $q<0$, then
A={1\over 2c} \sqrt{-q}.
\end{displaymath} (5)

Now use Partial Fraction Decomposition,
{1\over c} \int {du\over (u+A)(u-A)} = {1\over c} \int \left({{A_1\over u+A}+{A_2\over u-A}}\right)\,du
\end{displaymath} (6)

\left({{A_1\over u+A}+{A_2 \over u-A}}\right)= {A_1(u-A)+A_2(u+A)\over u^2-A^2} = {(A_1+A_2)u+A(A_2-A_1)\over u^2-A^2},
\end{displaymath} (7)

so $A_2+A_1=0 \Rightarrow A_2=-A_1$ and $A(A_2-A_1)=-2AA_1=1\Rightarrow A_1 = -1/(2A)$. Plugging these in,
${1\over c} \int \left({-{1\over 2A} {1\over u+A} + {1\over 2A} {1\over u-A}}\right)\, du$
$\quad = {1\over 2Ac} \left[{-\ln(u+A)+\ln(u-A)}\right]$
$\quad = {1\over 2Ac}\ln\left({u-A\over u+A}\right)$
$\quad = {1\over 2\left({1\over 2c}\right)\sqrt{-q}\,c} \ln\left({x+{b\over 2c}-{1\over 2c} \sqrt{-q}\over x+{b\over 2c}+{1\over 2c} \sqrt{-q}}\right)$
$\quad = {1\over\sqrt{-q}} \ln\left({2cx+b-\sqrt{-q}\over 2cx+b+\sqrt{-q}}\right)$ (8)
for $q<0$. Note that this integral is also tabulated in Gradshteyn and Ryzhik (1979, equation 2.172), where it is given with a sign flipped.


Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 5th ed. San Diego, CA: Academic Press, 1979.

© 1996-9 Eric W. Weisstein