Understanding a solution of $\sqrt{y^2+1}\,dx = xy\,dy$: where does the constant come from after integrating? They got
$$\frac{dx}{x} = \frac{y \, dy}{\sqrt{y^2+1}},$$
and in the next line
$$ \int \frac{dx}{x} = \int \frac{y \, dy}{\sqrt{y^2+1}} + C.$$
Where is the $C$ from?
 A: Summarizing the above comments:

As an indefinite integral has a constant (you should have learnt indefinite integrals before learning differential equations), consequently there must be a constant to cover all solutions for the differential equation.

A: Equaling two expressions in $dx$ and $dy$, actually make sense (implicitly means) that we are considering
$x$ and $y$ as  functions of a same variable, say $t$.
Where the $t$ can then be the $x$ or the $y$ itself.
That premised, we intend that
$$
dx = d\left( {x + c_{\,x} } \right)\;:dx(t) = d\left( {x(t) + c_{\,x} } \right)
$$
and same for $y$.
Therefore the expression
$$
{{d\xi } \over {\left( {\xi  - c_{\,x} } \right)}} = {{\left( {\eta  - c_{\,y} } \right)d\eta } \over {\sqrt {\left( {\eta  - c_{\,y} } \right)^{\,2}  + 1} }}
$$
shall be understood with $\xi , \, \eta$ as functions of $t$, and it shall be integrated over the same interval for the parameter, i.e.:
$$
\int_{t = t_{\,0} }^{\,t} {{{d\xi } \over {\left( {\xi  - c_{\,x} } \right)}}}  = \int_{t = t_{\,0} }^{\,t} {{{\left( {\eta  - c_{\,y} } \right)d\eta } \over {\sqrt {\left( {\eta  - c_{\,y} } \right)^{\,2}  + 1} }}} 
$$
Since the constants are arbitrary we can write
$$
\xi (t_{\,0} ) - c_{\,x}  = a\quad \xi (t) - c_{\,x}  = b(t)\quad \eta (t_{\,0} ) - c_{\,y}  = c\quad \eta (t) - c_{\,y}  = d(t)
$$
and therefore
$$
\int_{\xi  = a}^{\,b(t)} {{{d\xi } \over \xi }}  = \int_{\eta  = c}^{\,d(t)} {{{\eta d\eta } \over {\sqrt {\eta ^{\,2}  + 1} }}} 
$$
which means
$$
\int {{{d\xi } \over \xi }}  + C_{\,\xi }  = \int {{{\eta d\eta } \over {\sqrt {\eta ^{\,2}  + 1} }}}  + C_{\,\eta } 
$$
or
$$
\int {{{d\xi } \over \xi }}  = \int {{{\eta d\eta } \over {\sqrt {\eta ^{\,2}  + 1} }}}  + C
$$
It is clear that, in the general case, the range of $C$ will be subject to limitations.
A: Strictly speacking, the integration constant at that place is superfluous. The indefinite integrals already implicitly contain an indeterminate constant. So a better way is to have the constant only appear after actually selecting one of the anti-derivative functions,
$$
\int \frac{dx}x=\int\frac{y\,dy}{\sqrt{1+y^2}}\\~\\
\implies
\ln|x|=\ln\sqrt{1+y^2}+C.
$$
