I am going over Leibnitz's Rule, and unfortunately one part isn't making sense to me. Leibnitz's Rule states $$\frac {\partial}{\partial{y}}\int M(t, y)dt = \int \frac {\partial{M(t, y)}}{\partial{y}}dt$$
Specifically, suppose that $M(t, y) = t+y$.
Then, $\frac{\partial{M}}{\partial{t}} = 1$ and $\int 1dt = t+h(y)+c$.
However, $\int (t+y)dt = \frac{t^2}{2}+yt+h(y)+c$, and $\frac{\partial}{\partial{y}}(\frac{t^2}{2}+yt+h(y)+c) = t$. However, this would mean that $t+h(y)+c = t$ only when $h(y) +c = 0$. In other words, this would mean that $$\frac {\partial}{\partial{y}}\int M(t, y)dt = \int \frac {\partial{M(t, y)}}{\partial{y}}dt$$ is only true when the constant of integration and arbitrary function $h(y)$ are 0...?
I would really appreciate any and all clarification on this. Thanks so much.
Rather than use the same $h(y)+c$ for both integrations and worrying why they are different, let us use different letters, $g(y)$ for the integration of which we differentiate, and $h(y)$ for the integration of a differentiation, then compare them. (Note: The arbitrary constants may be folded into the arbitrary functions because they are arbitrary.)
$$\begin{align}\dfrac{\partial~~}{\partial y}\int y+t\,\mathrm d t &= \dfrac{\partial(yt+\tfrac 12 t^2+g(y))}{\partial y}\\[1ex]&=t+g'(y)\\[2ex]\int\dfrac{\partial (y+t)}{\partial y}\mathrm d t&=\int1\,\mathrm d t\\[1ex]&=t+h(y)\end{align}$$
Result: We have some arbitrary functions $g$ and $h$, of $y$, where it turns out that $h(y)=\dfrac{\mathrm d g(y)}{\mathrm d y}$.
Therefore, there is nothing alarming happening here.