There are two things I don't understand in this derivation. The first is that the total time derivative $\frac{d}{d t}$ turns into a partial time derivative $\frac{\partial}{\partial t}$ when inside the integral.
Generally speaking, this is implied by the <PERSON> intregral rule, which is a particular case of differentiation under the integral sign. For a function $f(x,t)$, the rule implies: $$ \frac{d}{dt} \left(\int_{x_0}^{x_1} f(x,t)dx \right) = \int_{x_0}^{x_1} \frac{\partial f(x,t)}{\partial t} dx$$ See the links for derivations, and at least note that if the limits of integration are themselves time dependent functions, their derivatives become important and you will have two additional terms outside the integral sign. The <PERSON> rule typically refers to the case where the limits are constants and so their derivatives (and hence the two extra terms) are zero.
The second is that the author states: $(\hat{H}\psi)^ = \psi^\hat{H}$. This confuses me as $\hat{H}$ is the Hamiltonian operator, and contains an $x$ derivative.
Going along with <PERSON> answer, you'll find that it's often more useful and illuminating to think about these expressions with your linear algebra goggles on -- that is, thinking about operators as self-adjoint matrices, rather than differential operators acting on functions in the Cartesian basis. 526
The latter gets real ugly, real fast when you get into 3-dimensional QM. Dirac (bra-ket) notation will be your best friend, I promise.
But now that we've stressed that enough, I suppose an example might be helpful in thinking about linear differential operators. Consider the 1-D momentum operator, $\hat{p} = \frac{\hbar}{i} \frac{d}{dx}$. Let's now show that this operator is Hermitian: $$ = \int_{-\infty}^{\infty} f^ \frac{\hbar}{i}\frac{dg}{dx}dx$$ Now, apply integration by parts. This gives: $$\int_{-\infty}^{\infty} f^ \frac{\hbar}{i}\frac{dg}{dx}dx = \frac{\hbar}{i}f^ g |{-\infty}^{\infty} + \int{-\infty}^{\infty} \left(\frac{\hbar}{i}\frac{df}{dx}\right)^ g dx = <\hat{p}f|g>$$ Note that the first term must be zero, given that wavefunctions in QM are square integrable, or "normalizable," which requires that they go to zero at $\pm \infty$.
[Example source: Griffiths Introduction to QM, section 3.2.1] 916