If slope of the tangent at the point (x, y) on the curve is `(y-1)/(x^(2)+x)`, then the equation of the curve passing through M(1, 0) is :

Options

A. `(y-1)(x+1)-2x=0`
B. `(y-1)(x-1)-2x=0`
C. `(y+1)(x+1)-2x=0`
D. `(y+1)(x-1)+2x=0`

Explanation

To solve the problem, we need to find the equation of the curve given the slope of the tangent at any point (x, y) on the curve, which is given by the expression \((y - 1)/(x^2 + x)\). The curve also passes through the point \(M(1, 0)\).

### Step-by-Step Solution:

1. **Set Up the Differential Equation:**
The slope of the tangent can be expressed as:
\[ \frac{dy}{dx} = \frac{y - 1}{x^2 + x} \]

2. **Separate the Variables:**
We can rearrange the equation to separate the variables \(y\) and \(x\):
\[ \frac{dy}{y - 1} = \frac{dx}{x^2 + x} \]

3. **Integrate Both Sides:**
Now, we will integrate both sides. The left-hand side requires the integral of \(\frac{1}{y - 1}\) and the right-hand side requires the integral of \(\frac{1}{x^2 + x}\).
- For the left side:
\[ \int \frac{dy}{y - 1} = \ln |y - 1| + C_1 \]
- For the right side, we can factor \(x^2 + x\) as \(x(x + 1)\) and use partial fractions:
\[ \frac{1}{x^2 + x} = \frac{1}{x} - \frac{1}{x + 1} \]
Thus, we integrate:
\[ \int \left( \frac{1}{x} - \frac{1}{x + 1} \right) dx = \ln |x| - \ln |x + 1| + C_2 \]

4. **Combine the Integrals:**
Combining both integrals, we have:
\[ \ln |y - 1| = \ln |x| - \ln |x + 1| + C \]
where \(C = C_2 - C_1\).

5. **Exponentiate to Remove Logarithms:**
By exponentiating both sides, we get:
\[ |y - 1| = K \cdot \frac{x}{x + 1} \]
where \(K = e^C\).

6. **Remove Absolute Values:**
Since we are looking for a function, we can drop the absolute value (assuming \(y\) is greater than 1):
\[ y - 1 = K \cdot \frac{x}{x + 1} \]

7. **Find the Constant \(K\) Using the Point \(M(1, 0)\):**
Substitute \(x = 1\) and \(y = 0\) into the equation:
\[ 0 - 1 = K \cdot \frac{1}{1 + 1} \Rightarrow -1 = K \cdot \frac{1}{2} \Rightarrow K = -2 \]
Therefore, we have:
\[ y - 1 = -2 \cdot \frac{x}{x + 1} \]

8. **Rearranging to Get the Final Equation:**
Rearranging gives:
\[ y - 1 = -\frac{2x}{x + 1} \Rightarrow (x + 1)(y - 1) = -2x \]
Thus, the equation of the curve is:
\[ (x + 1)(y - 1) + 2x = 0 \]

### Final Answer:
The equation of the curve is:
\[ (x + 1)(y - 1) - 2x = 0 \]