My favorite method for this is to use Gauss's theorem in reverse, so to speak. For brevity, let me write $\Delta u$ for the Laplacian and $u_r$, $u_\theta$ etc for the partial derivatives. Note that $\Delta u$ is the divergence of $\nabla u$, so Gauss's theorem says $ \iint_\Omega\Delta u\,dx\,dy=\int_{\partial\Omega}\mathbf{n}\cdot\nabla u\,ds .$ Apply this to the domain $\Omega$ given by $r_1 and $\theta_1<\theta<\theta_2$, and note that
- On the boundary $r=r_2$, $\mathbf{n}\cdot\nabla u=u_r$ and $ds=r_2\,d\theta$
- On the boundary $r=r_1$, $\mathbf{n}\cdot\nabla u=-u_r$ and $ds=r_1\,d\theta$
- On the boundary $\theta=\theta_2$, $\mathbf{n}\cdot\nabla u=u_\theta/r$ and $ds=dr$
- On the boundary $\theta=\theta_1$, $\mathbf{n}\cdot\nabla u=-u_\theta/r$ and $ds=dr$
so Gauss becomes $\begin{aligned}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}\Delta u\cdot r\,dr\,d\theta &=\int_{\theta_1}^{\theta_2}\bigl(r_2u_r(r_2,\theta)-r_1u_r(r_1,\theta)\bigr)\,d\theta\\ &\quad+\int_{r_1}^{r_2}\frac{u_\theta(r,\theta_2)-u_\theta(r,\theta_1)}{r}\,dr\\ &=\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(ru_r)_r\,dr\,d\theta +\int_{r_1}^{r_2}\int_{\theta_1}^{\theta_2}\frac{u_{\theta\theta}}{r}\,d\theta\,dr\\ &=\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}\Bigl((ru_r)_r+\frac{u_{\theta\theta}}{r}\Bigr)\,dr\,d\theta \end{aligned}$ Since this holds for all choices of the limits, the integrands must be the same, so $\Delta u\cdot r=(ru_r)_r+\frac{u_{\theta\theta}}{r}.$ Now divide by $r$.