I am starting the chapter on differential equations and have this example to work through but I do not understand a few things
Solve $dy=\frac{dy}{dx}=\frac{2x(y-1)}{x^2+1}$
solution:
note that $y(x)=1 $ is a solution (where does this come from?)
now assume that $y(x)\neq1$ and separate the variables
$\frac{dy}{y-1}= \frac{2x}{x^2+1}$ (this I understand)
$\int\frac{dy}{y-1}dy=\int \frac{2x}{x^2+1}dx$
$ln|y-1| = ln(x^2+1)+c$
why on LHS is there no $+C$ and why is the RHS $ln(x^2+1)$ and not $ln|x^2+1|$
To find the solution $y(x)=1$ just try setting both sides to $0$.
$x^2+1$ is strictly positive, so $\ln(x^2+1)=\ln|x^2+1|$.
Finally you can write the LHS as $\ln|y-1|+c_1$, and the RHS as $\ln(x^2+1)+c_2$. Then $c=c_2-c_1$. So only one side needs the constant.