Integral of partial derivative: $\int_0^\mu \frac{\partial q} {\partial u}\,du$

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An equation (borrowed from statistics QLE) is

$$ \frac{\partial q(\mu,y)} {\partial\mu}= \dfrac{y-\mu}{a_i \phi V(\mu)}$$

Apparently this is equivalent to

$$q(\mu,y)=\int_0^\mu \frac{y-u}{a_i \phi V(u)}\,du + \textrm{a function of }y.$$

I'm not 100% sure how they got this equation. It seems like they computed

$$\int_0^\mu \frac{\partial q(u,y)} {\partial u}\,du = q(\mu,y) + \textrm{a function of }y$$

from the LHS. Instinctively, I try to evaluate this integral by cancelling the differentials of $u$, but I'm not sure that is allowed because one is a $d$ and one is a $\partial$, and even then the integral would turn into (I believe) $\int_0^\mu \partial q=\mu$, which does not give the equivalent statement. Can someone help clear up my confusion? Thanks!