$

x = u+h \\

y = v+k\\

$

at $(u,v) = 0\\$

$

k - h - 2 = 0 \\

h + k = 0 \\

\Rightarrow h = -1,\;k = 1\\

$

$

x = u - 1 \\

dx = du\\

y = v + 1 \\

dy = dv \\

$

$$

\frac{dv}{du} = \frac{v-u}{u+v} \\

$$

let $v = ut,\;\frac{dv}{du}=t+u \frac{dt}{du}$

$$

t + u \frac{dt}{du}=\frac{ut-u}{u+ut} = \frac{t-1}{1+t}

$$

simplify with magic

$$

\frac{1}{u} du = \frac{1+t}{-1-t^2}dt \\

\ln \left| u \right| = -\frac{1}{2}\ln \left| t^2 +1 \right| -\arctan t + C \\

\ln \left| u \right| = -\frac{1}{2}\ln \left| \left( \frac{v}{u} \right) ^2 +1 \right| -\arctan \left( \frac{v}{u} \right) + C \\

\ln \left| x+1 \right| = -\frac{1}{2}\ln \left| \left( \frac{y-1}{x+1} \right) ^2 +1 \right| -\arctan \left( \frac{y-1}{x+1} \right) + C

$$