The solution of the equation $\frac{\mathrm{dy}}{\mathrm{d} x}=\frac{1}{x+\mathrm{y}+1}$ is

Substitute $z = x + y + 1$, so that $\frac{dz}{dx} = 1 + \frac{dy}{dx}$. Substituting into the differential equation $\frac{dy}{dx} = \frac{1}{z}$ yields

$\frac{dz}{dx} - 1 = \frac{1}{z}$

which rearranges to

$\frac{dz}{dx} = \frac{z + 1}{z}$

Separating variables gives

$\frac{z}{z+1} \, dz = dx$

Integrate both sides:

$\int \left(1 - \frac{1}{z+1}\right) dz = \int dx$

$z - \ln |z+1| = x + c$

Substituting back $z = x + y + 1$ produces

$(x + y + 1) - \ln |x + y + 2| = x + c$

Simplifying,

$y + 1 - \ln |x + y + 2| = c$

Rewriting,

$y = \ln |x + y + 2| + C$

where $C = c - 1$ is an arbitrary constant.
This matches option C.