The solution of $\frac{\mathrm{dy}}{\mathrm{d} x}=(x+\mathrm{y})^2$ is

Solve the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x} = (x + y)^2$ by substituting $z = x + y$.

Differentiating the substitution with respect to $x$ gives $\frac{\mathrm{d}z}{\mathrm{d}x} = 1 + \frac{\mathrm{d}y}{\mathrm{d}x}$, so $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}z}{\mathrm{d}x} - 1$.

Substituting into the original equation yields $\frac{\mathrm{d}z}{\mathrm{d}x} - 1 = z^2$, which rearranges to $\frac{\mathrm{d}z}{\mathrm{d}x} = 1 + z^2$.

Separate variables: $\frac{\mathrm{d}z}{1 + z^2} = \mathrm{d}x$, then integrate both sides: $\int \frac{\mathrm{d}z}{1 + z^2} = \int \mathrm{d}x$.

The left side is $\tan^{-1}(z)$, and the right side is $x + c$, where $c$ is the integration constant.

Thus, $\tan^{-1}(z) = x + c$, and substituting back $z = x + y$ gives $\tan^{-1}(x + y) = x + c$.

The solution corresponds to option A, so the final answer is $\boxed{\text{A}}$.