This answers the original version of the question:
Use $ \vert \vec{r} - c \hat{z} \vert^2 = r^2 + c^2 - 2 r c \cos \theta$. Now integration with respect to $\theta$ can be carried out, change $t = \cos\theta$.
$ \begin{eqnarray} 2 \pi \int_0^ \pi \frac{r(1-\cos \theta)}{ \left( r^2 + c^2 - 2 \, c \cdot r \cdot \cos \theta \right)^{3/2}} \sin \theta \, \mathrm{d} \theta &=& 2 \pi \int_{-1}^1 \frac{r (1 -t) }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \mathrm{d} t \\ &=& 2 \pi \frac{ c^2 + r^2 - \vert c^2 - r^2 \vert }{c^2 r ( c+ r)} \end{eqnarray} $
Integration with respect to $r$ is now trivial:
$ \begin{eqnarray} \int_0^\infty 2 \pi \frac{ c^2 + r^2 - \vert c^2 - r^2 \vert }{c^2 r ( c+ r)} \mathrm{d} r &=& \int_0^c 2 \pi \frac{ 2 r^2 }{c^2 r ( c+ r)} \mathrm{d} r + \int_c^\infty 2 \pi \frac{ 2 c^2 }{c^2 r ( c+ r)} \mathrm{d} r \\ & = & 2 \pi \frac{2 - 2 \log 2}{c} + 2 \pi \frac{2 \log 2}{c} = \frac{4 \pi}{c} \end{eqnarray} $
It now remains to show steps to integrate with respect to $t$:
$ \begin{eqnarray} \frac{r (1 -t) }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } &=& \frac{1}{c} \frac{ (r^2 + c^2 - 2 r c t) - (r-c)^2 }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \\ &=& \frac{1}{c} \frac{1}{\left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{1/2}} - \frac{1}{c} \frac{(r-c)^2 }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \end{eqnarray} $ Now integration with respect to $t$ can be done trivially using $\int ( a-b t)^{-c} \mathrm{d} t =\frac{ (a - b t)^{1-c}}{b (c-1)}$.
Modified version
$ \begin{eqnarray} 2 \pi \int_0^ \pi \frac{(r-c \cos \theta)}{ \left( r^2 + c^2 - 2 \, c \cdot r \cdot \cos \theta \right)^{3/2}} \sin \theta \, \mathrm{d} \theta &=& 2 \pi \int_{-1}^1 \frac{(r -c t) }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \mathrm{d} t \\ &=& 2 \pi \frac{ 1 - \operatorname{sign}(c - r ) }{r^2} \end{eqnarray} $
Integrating with respect to $r$:
$ \int_0^\infty 2 \pi \frac{ 1 - \operatorname{sign}(c - r ) }{r^2} \mathrm{d} r = \int_c^\infty 2 \pi \frac{ 2 }{r^2} \mathrm{d} r = \frac{4\pi}{c} $
The logic of carrying out integration with respect to $t$ is the same.