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(A) Given the differential equation $\frac{dy}{dx} = \frac{(x-1)+(y-1)}{(x-1)-(y-1)}$.Let $X = x-1$ and $Y = y-1$,then $\frac{dY}{dX} = \frac{X+Y}{X-Y

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$2 	an^{-1}\left(\frac{1}{k}ight) = \log_{e}(k^{2}+1)$

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(A) Given the differential equation $\frac{dy}{dx} = \frac{(x-1)+(y-1)}{(x-1)-(y-1)}$.Let $X = x-1$ and $Y = y-1$,then $\frac{dY}{dX} = \frac{X+Y}{X-Y}$.Dividing numerator and denominator by $X$,we get $\frac{dY}{dX} = \frac{1 + (Y/X)}{1 - (Y/X)}$.Let $Y = VX$,then $\frac{dY}{dX} = V + X\frac{dV}{dX}$.Substituting this,$V + X\frac{dV}{dX} = \frac{1+V}{1-V}$,so $X\frac{dV}{dX} = \frac{1+V}{1-V} - V = \frac{1+V-V+V^{2}}{1-V} = \frac{1+V^{2}}{1-V}$.Separating variables,$\int \frac{1-V}{1+V^{2}} dV = \int \frac{dX}{X}$.$\int \frac{1}{1+V^{2}} dV - \frac{1}{2} \int \frac{2V}{1+V^{2}} dV = \ln|X| + C$.$	an^{-1}(V) - \frac{1}{2} \ln(1+V^{2}) = \ln|X| + C$.Substituting $V = Y/X = \frac{y-1}{x-1}$,we get $	an^{-1}\left(\frac{y-1}{x-1}ight) - \frac{1}{2} \ln\left(1 + \frac{(y-1)^{2}}{(x-1)^{2}}ight) = \ln|x-1| + C$.Since it passes through $(2,1)$,$X=1, Y=0$: $	an^{-1}(0) - \frac{1}{2} \ln(1) = \ln(1) + C \implies C = 0$.For point $(k+1, 2)$,$X=k, Y=1$: $	an^{-1}\left(\frac{1}{k}ight) - \frac{1}{2} \ln\left(1 + \frac{1}{k^{2}}ight) = \ln(k)$.$	an^{-1}\left(\frac{1}{k}ight) - \frac{1}{2} \ln\left(\frac{k^{2}+1}{k^{2}}ight) = \ln(k)$.$	an^{-1}\left(\frac{1}{k}ight) = \ln(k) + \frac{1}{2} \ln\left(\frac{k^{2}+1}{k^{2}}ight) = \ln(k) + \ln\left(\sqrt{\frac{k^{2}+1}{k^{2}}}ight) = \ln\left(k \cdot \frac{\sqrt{k^{2}+1}}{k}ight) = \ln\sqrt{k^{2}+1}$.Multiplying by $2$,we get $2 	an^{-1}\left(\frac{1}{k}ight) = \ln(k^{2}+1)$.

If the solution curve of the differential equation $\frac{dy}{dx} = \frac{x+y-2}{x-y}$ passes through the points $(2,1)$ and $(k+1, 2)$,where $k > 0$,then:

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