The electric field at a distance $r$ from am infinite line uniformly charged is
$$\vec{E}=\frac{\lambda}{2\pi r}\hat r.$$
This can be shown from Gauss's law.
If we want a potential whose negative gradient is $\vec{E}$, then clearly $V(r)=-\frac{\lambda}{2\pi}\log(r)$ will do the job.
However, if I wanted to obtain this potential directly from Coulomb's law and the superposition principle, I would write the integral
$$\int_{-\infty}^\infty \frac{\lambda dx}{\sqrt{r^2+x^2}},$$
and this integral is divergent.
So the question is how to obtain the potential from Coulomb's law and the superposition principle, without computing the electric field first.
Best Answer
You can do this on condition of not imposing a zero potential at infinity. If you impose a zero potential on the distance $r_0$ of the wire, the elementary potential will be of the form $dV=\frac{\lambda}{4\pi\epsilon}\left(\frac{1}{\sqrt{x^2+r^2}}-\frac{1}{\sqrt{x^2+{r_0}^2}}\right)$
It remains to integrate $\int_{-A}^{B}\left(\frac{1}{\sqrt{x^2+r^2}}-\frac{1}{\sqrt{x^2+{r_0}^2}}\right)dx=\hbox{arcsinh}\left(\frac{B}{r}\right)-\hbox{arcsinh}\left(\frac{B}{r_0}\right)+\hbox{arcsinh}\left(\frac{A}{r}\right)-\hbox{arcsinh}\left(\frac{A}{r_0}\right)$
Then to pass to the limit when $A$ and $B$ tend towards infinity $\hbox{arcsinh}\left(\frac{B}{r}\right)-\hbox{arcsinh}\left(\frac{B}{r_0}\right)\ -> \log(\frac{r_0}{r})$
We do indeed find the famous $2\log(\frac{r_0}{r})$ !