# Integral of $\frac{1}{\sqrt{(z-z')^2 + s^2}}$

I have a question about the signs of the antiderivative when one integrate $$\frac{1}{\sqrt{(z-z')^2 + s^2}}$$.

According to Wolfram Alpha here and here:

If one evaluates $$\int \frac{1}{\sqrt{(z-z')^2 + s^2}} dz$$, they get $$\log (z-z' + \sqrt{(z-z')^2 + s^2}) + C$$.

Evalutating $$\int \frac{1}{\sqrt{(z-z')^2 + s^2}} dz'$$ gives $$-\log (z-z' + \sqrt{(z-z')^2 + s^2}) + C$$.

However, one could rearrange the integrand before evaluating the second expression. Using the fact that $$(z-z')^2 = (z'-z)^2$$, one gets

$$$$\int \frac{1}{\sqrt{(z-z')^2 + s^2}} dz'= \int \frac{1}{\sqrt{(z'-z)^2 + s^2}} dz'$$$$

which has the same form as the first integral, implying that the two should be equal.

What am I missing here?

Notice that \begin{align} -\log{(z-z'+\sqrt{(z-z')^2+s^2})} &= \log{\left(\frac{1}{\sqrt{(z-z')^2+s^2} -(z'-z)}\right)} \\ &= \log{\left(\frac{\sqrt{(z-z')^2+s^2} +(z'-z)}{(\sqrt{(z-z')^2+s^2} -(z'-z))(\sqrt{(z-z')^2+s^2} +(z'-z)}\right)} \\ &= \log{\left(\frac{\sqrt{(z-z')^2+s^2} +(z'-z)}{s^2}\right)} \\ &= \log{(\sqrt{(z-z')^2+s^2} +z'-z)} -\log{s^2} \\ \end{align} So the two are equal except for a constant factor $$\log{s^2}$$