The equation of a curve passing through $(1,0)$ and having slope of tangent at any point $(\mathrm{x}, \mathrm{y})$ of the curve as $\frac{\mathrm{y}-1}{x^2+x}$ is

The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{y-1}{x^2+x}$.

Separating variables yields $\frac{dy}{y-1} = \frac{dx}{x(x+1)}$.

Integrating both sides gives $\int \frac{dy}{y-1} = \int \frac{dx}{x(x+1)}$.

The left side integrates to $\ln|y-1| + C_1$.

For the right side, decompose into partial fractions: $\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$.

Integrating gives $\ln|x| - \ln|x+1| + C_2 = \ln\left|\frac{x}{x+1}\right| + C_2$.

Combining results, $\ln|y-1| = \ln\left|\frac{x}{x+1}\right| + C$.

Using the point $(1, 0)$ to find $C$: $\ln|0-1| = \ln\left|\frac{1}{2}\right| + C$, so $0 = \ln\frac{1}{2} + C$ and $C = \ln2$.

Substituting back: $\ln|y-1| = \ln\left|\frac{x}{x+1}\right| + \ln2 = \ln\left|\frac{2x}{x+1}\right|$.

Exponentiating both sides: $|y-1| = \left|\frac{2x}{x+1}\right|$.

At $(1, 0)$, $y-1 = -1$ and $\frac{2x}{x+1} = 1$, indicating $y-1 = -\frac{2x}{x+1}$.

Multiplying both sides by $x+1$: $(y-1)(x+1) = -2x$.

Rewriting: $2x + (y-1)(x+1) = 0$, which corresponds to option C.
Final answer: $\boxed{\text{C}}$