Convergence of the integral Taking $b,c \in \mathbb{R^+}$, the integral $$\int_{-\infty}^{\infty} \dfrac{\mathrm{d}x}{\sqrt{\left(a-x\right)^2+b}}$$ 
diverges, however the integral 
$$\int_{-\infty}^{\infty} \left(\dfrac{1}{\sqrt{\left(a-x\right)^2+b}}-\dfrac{1}{\sqrt{\left(a-x\right)^2+c}}\right)\mathrm{d}x$$ 
converges to 
$\ln\left(c\right)-\ln\left(b\right)$
The analitical result of the first and second integrals without limits of integration are respectively
$$\ln\left(\left|\sqrt{\left(x-a\right)^2+b}+x-a\right|\right)+C
$$
$$\ln\left(\left|\sqrt{\left(x-a\right)^2+b}+x-a\right|\right)-\ln\left(\left|\sqrt{\left(x-a\right)^2+c}+x-a\right|\right)+C
$$
I haven't succeeded on evaluating the limit of the second expression, is there any trick?

As Jack D'Aurizio and Messney said in their comments, we can eliminate the constant $a$ by renaming our variable $x-a \rightarrow x$, leaving the evaluation of our integral as
$$\left. \ln\left(\left|\frac{\sqrt{\left(\frac{x}{\sqrt{b}}\right)^2+1}+\frac{x}{\sqrt{b}}}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2+1}+\frac{x}{\sqrt{c}}}\right|\right)\right|_{-\infty}^{+\infty}$$ 
wich equals
$$\lim_{x \to \infty} \ln\left(\left|\frac{\sqrt{\left(\frac{x}{\sqrt{b}}\right)^2+1}+\frac{x}{\sqrt{b}}}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2+1}+\frac{x}{\sqrt{c}}}\right|\right)-\ln\left(\left|\frac{\sqrt{\left(\frac{x}{\sqrt{b}}\right)^2+1}-\frac{x}{\sqrt{b}}}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2+1}-\frac{x}{\sqrt{c}}}\right|\right)$$ 
Dividing the numerator and denominator by $x$ on the first term, inserting the limit inside the logarith and multiplying and dividing by the conjugate of the numerator and denominator of the second we get
$$ \ln\left(\lim_{x \to \infty}\left|\frac{\sqrt{\left(\frac{1}{\sqrt{b}}\right)^2+\frac{1}{x^{2}}}+\frac{1}{\sqrt{b}}}{\sqrt{\left(\frac{1}{\sqrt{c}}\right)^2+\frac{1}{x^{2}}}+\frac{1}{\sqrt{c}}}\right|\right)-
\ln \left(\frac{1}{1}\right)+\lim_{x \to \infty} \ln\left(\left|\frac{\sqrt{\left(\frac{x}{\sqrt{b}}\right)^2+1}+\frac{x}{\sqrt{b}}}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2+1}+\frac{x}{\sqrt{c}}}\right|\right)$$ 
The second terrm is zero while the first and third terms are equal and both result in
$$\ln\left(\sqrt{\frac{c}{b}}\right) = \frac{1}{2}\ln\left(\frac{c}{b}\right) $$
So their sum is
$$\ln\left(\frac{c}{b}\right)=\ln(c)-\ln(b) $$
