We have,

$f^{\text{'}}\left(x\right)=\ln \left(x^2+1\right)$

$f\left(x\right)=\int \ln \left(x^2-1\right)dx$

$f\left(x\right)=x\ln \left(x^2-1\right)-\int \frac{2x^2}{x^2-1}dx$

$f\left(x\right)=x\ln \left(x^2-1\right)-2\int \left(\frac{x^2-1}{x^2-1}+\frac{1}{x^2-1}\right)dx$

$f\left(x\right)=x\ln \left(x^2-1\right)-2x-\ln \left(\frac{x-1}{x+1}\right)+C$

$f\left(2\right)=2\ln \left(3\right)-4-\ln \left(\frac{1}{3}\right)+C=0$ $\left[\because f\left(2\right)=0\right]$

$\Rightarrow C=4-3\ln 3$

$\therefore f\left(x\right)=x\ln \left(x^2-1\right)-2n\ln \frac{\left(x-1\right)}{\left(x+1\right)}+4-3\ln 3$

defined for $S$ infinite $C$ values possible in set $S$ such that $f^{\text{'}}\left(x\right)=\ln \left(x^2-1\right)$