$\begin{aligned} & \frac{d y}{d x}=\frac{y+1}{x^2-x} \\ & \therefore \int \frac{d y}{y+1}=\int \frac{d x}{x(x-1)} \\ & \therefore \int \frac{d y}{y+1}=\int\left[\frac{1}{x-1}-\frac{1}{x}\right] d x \Rightarrow \log |y+1|=\log |x-1|-\log |x|+ \\ & \log c \\ & \text { We have } x=2 \text { and } y=1 \\ & \therefore \log |2|=\log |1|-\log |1|+\log c \Rightarrow c=2 \\ & \therefore \log |y+1|=\log |x-1|-\log |x|+\log |2| \\ & \log |y+1|=\log \left|\frac{2(x-1)}{x}\right| \\ & \therefore y+1=\frac{2(x-1)}{x} \Rightarrow x y+x=2 x-2 \Rightarrow x y=x-2\end{aligned}$