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`

Answer

To find the equation of the curve given the slope of the tangent at a point (x, y) on the curve, we will follow these steps:

### Step 1: Set up the differential equation
The slope of the tangent is given by:
\[ \frac{dy}{dx} = \frac{y - 1}{x^2 + x} \]

### Step 2: Rearrange the equation
We can rearrange this equation to separate the variables:
\[ \frac{dy}{y - 1} = \frac{dx}{x^2 + x} \]

### Step 3: Integrate both sides
Now, we will integrate both sides:
\[ \int \frac{dy}{y - 1} = \int \frac{dx}{x^2 + x} \]

The left side integrates to:
\[ \ln |y - 1| \]

For the right side, we can factor the denominator:
\[ x^2 + x = x(x + 1) \]
We can use partial fraction decomposition:
\[ \frac{1}{x(x + 1)} = \frac{A}{x} + \frac{B}{x + 1} \]
Multiplying through by the denominator \(x(x + 1)\) gives:
\[ 1 = A(x + 1) + Bx \]
Setting \(x = 0\) gives \(A = 1\), and setting \(x = -1\) gives \(B = -1\). Thus:
\[ \frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1} \]
Now we integrate:
\[ \int \left( \frac{1}{x} - \frac{1}{x + 1} \right) dx = \ln |x| - \ln |x + 1| = \ln \left| \frac{x}{x + 1} \right| \]

### Step 4: Combine the results
Combining both integrals, we have:
\[ \ln |y - 1| = \ln \left| \frac{x}{x + 1} \right| + C \]

### Step 5: Exponentiate both sides
Exponentiating both sides gives:
\[ |y - 1| = k \cdot \frac{x}{x + 1} \]
where \(k = e^C\).

### Step 6: Remove the absolute value
We can drop the absolute value (assuming \(y - 1\) is positive for the curve):
\[ y - 1 = k \cdot \frac{x}{x + 1} \]
Thus:
\[ y = k \cdot \frac{x}{x + 1} + 1 \]

### Step 7: Use the point M(1, 0) to find k
Substituting the point \(M(1, 0)\):
\[ 0 = k \cdot \frac{1}{1 + 1} + 1 \]
This simplifies to:
\[ 0 = \frac{k}{2} + 1 \implies k = -2 \]

### Step 8: Substitute k back into the equation
Substituting \(k\) back, we get:
\[ y = -2 \cdot \frac{x}{x + 1} + 1 \]
This simplifies to:
\[ y = -\frac{2x}{x + 1} + 1 = 1 - \frac{2x}{x + 1} = \frac{x + 1 - 2x}{x + 1} = \frac{1 - x}{x + 1} \]

### Final equation of the curve
Thus, the equation of the curve is:
\[ y = \frac{1 - x}{x + 1} \]