Integrating Laplacian

Question

I came across the following result in the Griffith's textbook, $$\int \frac{d^2\psi}{dr^2}dr=\Delta \psi',$$ where $\psi(r)=rR(r)$ is a solution to the radial Schrödinger equation $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dr^2}+\left(V(r)+\frac{\hbar}{2m}\frac{l(l+1)}{r^2}\right)\psi=E\psi.$$ However, mathematically, I don't see how this is possible, why would integral of the Laplacian of the function equal the Laplacian of the derivative?