Computing $\int\frac {du}{\sqrt{u^2 + s^2}} =\log \lvert (u + \sqrt{u^2 + s^2}) \rvert$ with a substitution Can someone please show me where I am going wrong? It seems there is a contradiction in the formula for the the following integral. 
$$\int\frac {du}{\sqrt{u^2 + s^2}} = \log \bigl\lvert u + \sqrt{u^2 + s^2}\bigr\rvert$$
Now, can't we rewrite the integral as follows using the substitution z = -u?
\begin{align}
\int\frac {du}{\sqrt{u^2 + s^2}} &= \int\frac {du}{\sqrt{(-u)^2 + s^2}} = \int\frac {-dz}{\sqrt{z^2 + s^2}}\\
& = -\log \bigl\lvert z + \sqrt{z^2 + s^2}\bigr\rvert = -\log \bigl\lvert (-u + \sqrt{u^2 + s^2})\bigr\rvert
\end{align}
But, clearly $\;\log \bigl\lvert u + \sqrt{u^2 + s^2}\bigr\rvert \ne -\log \bigl\lvert -u + \sqrt{u^2 + s^2} \bigr\rvert$
Also, the quantity inside $\log \bigl\lvert u + \sqrt{u^2 + s^2})\bigr\rvert$ is clearly non-negative for any choice of $u$ and $s$, so why do we need the absolute value of the quantity in our integral? 

You have a faster way, using some hyperbolic trigonometry:
Set $t=\dfrac us$, so that $\mathrm du=s\,\mathrm dt$. The integral becomes, supposing $s>0\,$:
$$\int\frac {\mathrm du}{\sqrt{u^2 + s^2}}=\int\frac {\not\! s\,\mathrm dt}{\not \!s\sqrt{ t^2 + 1}}=\int\frac {\mathrm dt}{\sqrt{ t^2 + 1}}=\operatorname{argsinh}t$$
and it is known that
$$\operatorname{argsinh}t =\ln\bigl(t+\sqrt{t^2+1}\bigr). $$
