<s>[INST] Integral table contradiction? In a table of integrals, I see the following two formulas:
$\int \frac{dx}{(a+x)(b+x)} = \frac{1}{b-a}\ln\frac{a+x}{b+x}$, and
$\int \frac{dx}{ax^2+bx+c} = \frac{2}{\sqrt{4ac-b^2}}\tan^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}}$.
How can these both be true?  It seems like if we expand $(a+x)(b+x)$ out to $x^2+(a+b)x+ab$, we can apply the 2nd equation to get
$\int \frac{dx}{(a+x)(b+x)} = \frac{2}{\sqrt{4ab-(a+b)^2}}\tan^{-1}\frac{2x+a+b}{\sqrt{4ab-(a+b)^2}}$,
which is surely not equivalent to $\frac{1}{b-a}\ln\frac{a+x}{b+x}$ (one involves a logarithm and the other involves an arctan, so no amount of algebraic fussing can reconcile them, can it?!)
 [/INST] Take a look at the logarithmic form of the complex arctangent:
$$\tan^{-1}x=\frac12i\Big(\ln(1-ix)-\ln(1+ix)\Big)\;.\tag{1}$$
For instance, consider the integral
$$\int\frac{dx}{(1+x)(2+x)}=\int\frac{dx}{x^2+3x+2}\;.$$ The two integration formulas yield the antiderivatives
$$\ln\frac{1+x}{2+x}\tag{2}$$ and
$$\frac2i\tan^{-1}\frac{2x+3}i\tag{3}\;.$$
From $(1)$ we have 
$$\begin{align*}
\frac2i\tan^{-1}\frac{2x+3}i&=\Big(\ln(1-(2x+3))-\ln(1+2x+3)\Big)\\
&=\ln\frac{-2-2x}{4+2x}\\
&=\ln\frac{-1-x}{2+x}\\
&=\ln\frac{1+x}{2+x}+\ln(-1)\\
&=\ln\frac{1+x}{2+x}+i\pi\;,
\end{align*}$$
using the principal value of the log. Thus, $(2)$ and $(3)$ really do differ only by a constant, and there is no problem.
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