In my text book there is an example of PDE with change of variable like this: $$\frac{\partial^2f}{\partial t^2}-c^2\frac{\partial ^2f}{\partial x^2}=0$$ with $\left\{\begin{matrix} u=x+ct & \\ v=x-ct & \end{matrix}\right.$
I can follow my text book until here $\frac{\partial f}{\partial v} = g(v)$ and then my book suggest that we should integrate both side with respect to '$v$' but it doesn't explain why the result become this $f(u,v)=G(v)+H(u)$ where G Is antiderivative of g and H is an arbitrary function of one variable but when I integrate I get something like this $$\int \frac{\partial f}{\partial v} dv = F(u,v) + H(u)$$ $$\int g(v) dv = G(v) + C $$ $$F(u,v) + H(u)=G(v) + C$$ $$ F(u,v)= G(v) - H(u) + C$$ i wonder why in my text book wrote like that: $F(u,v) = G(v) + H(u)$ ? Can someone please explain to me, I must say that i am a calc 2 student and don't have much experience in PDE.
Recall that the function $f$ depends on the two variables $u$ and $v$.
Integrating $g(v)$ with the absence of $u$ would lead to $G(v) + C$, where $C$ is a real constant and $G$ is an antiderivative of $g$.
But since you here also have the $u$, the $C$ is not enough. $C$ has to be replaced by an arbitrary function $H(u)$, because taking the partial derivative of $H(u)$ with respect to $v$ yields $0$. Of course, in $H(u)$ all constants $C$ are included as a special case. Hence
$$\int g(v) \ \mathrm dv = G(v) + H(u). \tag{1}$$
Your approach with
$$F(u, v) = G(v) - H(u) + C \tag{2}$$
is equivalent to this. In (2), include the $C$ in $H(u)$ and replace $-H(u)$ with $H(u)$ and you obtain (1).