Canceling out integral Bear with my naivety, I wanted to ask if it is possible to cancel out $\int$ with a $\frac{\mathrm d}{\mathrm dx}$. I had $\frac{\partial}{\partial v}$ in a question and I took $\partial v$ to the other side and took integration on both sides. So now on left hand side only $\partial$ remains so will it cancel out with the integral?
 A: Usually when we take integration we remove d or $\partial$
You can say it is cancel out or it is integrated.
Example -
$\frac{d}{dv} x = v^2 + 3$
$dx =  (v^2 + 3) dv$
On integrating both sides,
x = $\int(v^2 + 3)dv$
A: I wouldn't say that $\frac{\mathrm d}{\mathrm dx}$ and $\int$ cancel each other out. Consider the following examples
$$\frac{\mathrm d}{\mathrm dx}\int e^t\ \mathrm dt=\frac{\mathrm d}{\mathrm dx}\left(e^t+C\right)=0$$
$$\frac{\mathrm d}{\mathrm dx}\int_0^2 e^x\ \mathrm dx=\frac{\mathrm d}{\mathrm dx}\left(e^2-1\right)=0$$
Study the fundamental theorem of calculus for a deeper understanding of what's going on.

I'll also address what you mentioned in the comments. Although $\frac{\mathrm dM}{\mathrm dx}$ look and act like a ratio, in modern mathematical analysis, it's not a ratio. It just so happens that
$$\mathrm dM=\frac{\mathrm dM}{\mathrm dx}\ \mathrm dx$$
Therefore
$$V=\frac{\mathrm dM}{\mathrm dx}$$
$$\int V\ \mathrm dx=\int\frac{\mathrm dM}{\mathrm dx}\ \mathrm dx=\int\ \mathrm dM=M+C$$
