$\int\frac{1}{y^2-1}dy=\int\frac{1}{x}dx$

$\frac{1}{y^2-1}=\frac{A}{y+1}+\frac{B}{y-1}$

$Ay-A+By+B=1$

$A=-\frac{1}{2}, B=\frac{1}{2}$

$\int\Big(\frac{1}{-2(y+1)}+\frac{1}{2(y-1)}\Big)dy=\int\frac{1}{x}dx$

$-\frac{1}{2}\ln|y+1|+\frac{1}{2}\ln|y-1|=\ln|x|+C$

$\ln\Bigg|\frac{y-1}{y+1}\Bigg|=2\ln|x|+C$

$\frac{y-1}{y+1}=Cx^2$

Use$x = 1$ and $y = 2$ to solve for $C$.
Then use the value of $C$ and algebra to write the solution in $y =$ form.