The following is exercise 8.2 in 3rd edition (and exercise 8.19 in 2nd edition) of Goldstein's Classical Mechanics.
Adding the total time derivative of a function of $q_i$ and t to the Lagrangian will not change the the Euler-Lagrangian equation. So if we make the following change to Lagrangian,
$$L'(q,\dot{q},t) = L(q,\dot{q},t) + \frac{dF(q_1,q_2,…,q_n,t)}{dt}$$
we can get
$$
\frac{d}{dt}\frac{\partial{L'}}{\partial{\dot{q_i}}} – \frac{\partial{L'}}{\partial{q_i}} = 0
$$
from
$$
\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q_i}}} – \frac{\partial{L}}{\partial{q_i}} = 0
$$
How can we get the corresponding Hamiltonian equation part? This is to prove
$$
\dot{p'_i} = \frac{\partial{H'}}{\partial q_i}
$$
$$
-\dot{q_i} = \frac{\partial{H'}}{\partial p'_i}
$$
from
$$
\dot{p_i} = \frac{\partial{H}}{\partial q_i}
$$
$$
-\dot{q_i} = \frac{\partial{H}}{\partial p_i}
$$
where $p'_i = \frac{\partial L'}{\partial \dot q_i}$.
Edit
The corresponding $H'$ is
$$
H' = \sum_k{p'_k \dot{q_k}} – L'
$$
where $p'_k = \frac{\partial L'}{\partial \dot q_k}$.
Best Answer
If $$ L \to L' = L +\frac{dF(q,t)}{dt}$$ the corresponding Hamiltonian becomes $$ H \to H' = H - \frac{\partial F(q,t)}{\partial t} $$ as shown here. Moreover, the canonical momentum becomes $$ p \to P = p + \frac{\partial F}{\partial q} $$ while $$ q \to Q = q $$ as shown here.
These formulas allow us to check the invariance of Hamilton's equations explicitly. Concretely, \begin{align} \frac{dq}{dt} &= \frac{\partial H}{\partial p} \notag \\ \end{align} becomes \begin{align} \frac{dQ}{dt} &= \frac{\partial H'}{\partial P} \\ \therefore \quad \frac{dq}{dt} &= \frac{\partial \left(H - \frac{\partial F(q,t)}{\partial t} \right)}{\partial P} \\ \therefore \quad \frac{dq}{dt} &= \frac{\partial H}{\partial P} - \frac{\partial }{\partial P} \left( \frac{\partial F(q,t)}{\partial t} \right) \\ \therefore \quad \frac{dq}{dt} &= \frac{\partial H}{\partial P} \\ \therefore \quad \frac{dq}{dt} &= \frac{\partial H}{\partial p} \frac{\partial P}{\partial p} \\ \therefore \quad \frac{dq}{dt} &= \frac{\partial H}{\partial p} \quad \checkmark \end{align} where I used that $F$ does not depend on $P$ and $$\frac{\partial P}{\partial p} =\frac{\partial }{\partial p} \left( p+ \frac{\partial F}{\partial q} \right) = 1. $$
Analogously, we can check Hamilton's second equation: $$ \frac{dp}{dt}= -\frac{\partial H(q,p,t)}{\partial q} .$$ However, there is a subtlety. After the transformation, we have on the right-hand side $\frac{\partial H'(Q,P,t)}{\partial Q}$. But here we need take into account that $p$ also depends on $q$, since $ p \to P = p + \frac{\partial F(Q,t)}{\partial Q} $. Therefore \begin{align} \frac{\partial H'(Q,P,t)}{\partial Q} &= \frac{\partial H'(Q,p + \frac{\partial F}{\partial q} ,t)}{\partial Q} \\ &= \frac{\partial H'(Q,p,t)}{\partial Q} + \frac{\partial H(Q,p,t) }{\partial p} \frac{\partial p}{\partial Q} \\ &= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{\partial^2 F(Q,t)}{\partial Q \partial t} + \frac{\partial H(Q,p,t) }{\partial p} \frac{\partial p}{\partial Q} \\ &= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{\partial^2 F(Q,t)}{\partial Q \partial t} + \dot Q \frac{\partial \left(P- \frac{\partial F}{\partial q} \right)}{\partial Q} \\ &= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{\partial^2 F(Q,t)}{\partial Q \partial t} - \dot Q \frac{\partial }{\partial Q} \frac{\partial F}{\partial q} \\ &= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{d}{dt} \frac{\partial F}{\partial Q} \,. \end{align} where we used that $$ \frac{d}{dt} \frac{\partial F}{\partial Q}= \frac{\partial^2 F(Q,t)}{\partial Q \partial t} + \dot Q \frac{\partial }{\partial Q} \frac{\partial F}{\partial q} . $$
Using this, we can rewrite Hamilton's second equation after the transformation as follows: \begin{align} \frac{dP}{dt}&= -\frac{\partial H'(Q,P,t)}{\partial Q} \\ \therefore \quad \frac{d}{dt} \left( p+ \frac{\partial F(q,t)}{\partial q} \right) &= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{d}{dt} \frac{\partial F}{\partial Q} \\ \therefore \quad \frac{dp}{dt} + \frac{d}{dt} \left(\frac{\partial F(q,t)}{\partial q} \right)&= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{d}{dt} \frac{\partial F}{\partial Q} \\ \therefore \quad \frac{dp}{dt} &= -\frac{\partial H}{\partial q} \quad \checkmark \end{align}
EDIT: The subtlety was also noted here, but unfortunately without an answer and a few years ago there was even a paper which didn't notice it and claimed that Hamilton's equations are not invariant.