Integral of Rational Functions $$\int \frac{dx}{ax^2 + bx + c} \quad \text{for} \quad 4ac-b^2 >0$$
then
$$\begin{align}
ax^2 + bx + c
&= a\biggl(x+\frac{b}{2a}\biggr)^2 + \frac{4ac-b^2}{4a} \\
&= \Biggl(\sqrt{a}\biggl(x+\frac{b}{2a}\biggr)\Biggr)^2 + \Biggl(\sqrt{\frac{4ac-b^2}{4a}}\Biggr)^2
\end{align}$$
implies
$$ \int \frac{dx}{ax^2 + bx + c} = \int \frac{dx}{\Bigl(\sqrt{a}\bigl(x+\frac{b}{2a}\bigr)\Bigr)^2 + \Bigr(\sqrt{\frac{4ac-b^2}{4a}}\Bigr)^2} = \frac{2\sqrt{a}}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}}
$$
The correct answer does not have the $\sqrt{a}$ in the numerator.  I've checked over my work, but have clearly made some dumb mistake.  Can anyone please let me know where such an error was made?  Thank you for your help!

Hard to debug, since some detail is missing. To make things less messy, I would rewrite the integral as
$$ \int\frac{4a\,dx}{(2ax+b)^2+4ac-b^2}.$$
Let $4ac-b^2=K^2$.  Then let $2ax+b=Ku$. We have $2a \,dx=K\,du$. So our integral is 
$$\int \frac{2K\,du}{K^2u^2+K^2},$$
which is 
$$\frac{2}{K}\arctan u+C.$$
Remark: I am not fond of fractions, so for completing the square I prefer to write $$ax^2+bx+c=\frac{1}{4a}\left(4a^2x^2+4abx+4c\right)=\frac{1}{4a}\left((2ax+b)^2-(b^2-4ac)  \right).$$
