Hamiltonian density from Klein-Gordon field

In the solution for Peskin & Shroeder 2.2 where the Hamiltonian density obtained from the Klein-Gordon Lagrangian is given by: $$H = \pi^* \pi + \nabla \phi \cdot \nabla \phi^* + m^2 \phi^* \phi \\ = \pi^* \pi + \phi ^ * (- \nabla ^ 2 + m^2 ) \phi$$

Im confused with how the last equality was obtained. The notes said they used integration by parts and canceling out the surface term. Using the product rule $$\nabla \phi \cdot \nabla \phi^* = \frac{1}{2} (\nabla ^2(\phi \phi^*) - \phi \nabla^2 \phi^* - \phi^* \nabla^2 \phi)$$

Besides the surface term $$\nabla^2(\phi \phi^*)$$, I have an extra term which does not seem to evaluate to zero here

So, part (a) of problem 2.2 in P&S asks you to find the expression of the Hamiltonian (which you already found):$$H=\int d^3x \,\mathcal{H}=\int d^3x\,(\pi^*\pi+\nabla\phi^*\cdot\nabla\phi+m^2\phi^*\phi).$$To get to the next expression you mentioned, we use the chain rule (Einstein summation convention is assumed)$$\nabla\phi^*\cdot\nabla\phi=(\partial_i\phi^*)(\partial_i\phi)=\partial_i(\phi^*\partial_i\phi)-\phi^*(\nabla^2\phi)$$Now, substituting this expression in the Hamiltonian, we find that\begin{align*}H=&\int d^3x\,(\pi^*\pi+(\partial_i(\phi^*\partial_i\phi)-\phi^*(\nabla^2\phi))+m^2\phi^*\phi)\\ =&\int d^3x\,(\pi^*\pi-\phi^*(\nabla^2\phi)+m^2\phi^*\phi)+\int d^3x\;\partial_i(\phi^*\partial_i\phi)\\ =&\int d^3x\,(\pi^*\pi-\phi^*(\nabla^2\phi)+m^2\phi^*\phi)\end{align*}Note that the last term in step two is a surface term whose integral goes to zero. The integral of the surface term goes to zero because $$\phi,\phi^*\rightarrow0$$ as $$x\rightarrow\infty$$ which is a necessary condition for integrals like $$\int_{\text{all spacetime}} d^3x\,\phi^*\phi$$ (which are clearly present in the Hamiltonian and the Lagrangian) to make sense.
Therefore you have the desired expression for the Hamiltonian density:$$\mathcal{H}=\pi^*\pi+\phi^*(-\nabla^2+m^2)\phi$$