The equation of the curve through the point (1,0) whose slope is `(y-1)/(x^(2)+x)` is:

Options

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

Answer

To solve the problem of finding the equation of the curve through the point (1,0) whose slope is given by \(\frac{y-1}{x^2+x}\), we will follow these steps:

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

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

### Step 3: Integrate both sides
Now we will integrate both sides. The left-hand side becomes:
\[ \int \frac{dy}{y-1} = \ln|y-1| + C_1 \]
For the right-hand side, we will use partial fraction decomposition:
\[ \frac{1}{x^2+x} = \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \]
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:
\[ \frac{1}{x^2+x} = \frac{1}{x} - \frac{1}{x+1} \]
Now we can integrate:
\[ \int \left(\frac{1}{x} - \frac{1}{x+1}\right) dx = \ln|x| - \ln|x+1| + C_2 \]

### Step 4: Combine the results
Combining the results from both integrals, we have:
\[ \ln|y-1| = \ln|x| - \ln|x+1| + C \]
Where \(C = C_2 - C_1\).

### Step 5: Simplify the equation
Using properties of logarithms, we can rewrite this as:
\[ \ln|y-1| = \ln\left(\frac{x}{x+1}\right) + C \]
Exponentiating both sides gives:
\[ |y-1| = e^C \cdot \frac{x}{x+1} \]
Let \(k = e^C\), then:
\[ y - 1 = k \cdot \frac{x}{x+1} \]

### Step 6: Solve for \(y\)
Thus, we can express \(y\) as:
\[ y = k \cdot \frac{x}{x+1} + 1 \]

### Step 7: Determine the constant \(k\) using the initial condition
We know the curve passes through the point (1,0):
\[ 0 = k \cdot \frac{1}{1+1} + 1 \]
This simplifies to:
\[ 0 = \frac{k}{2} + 1 \implies k = -2 \]

### Step 8: Write the final equation
Substituting \(k\) back into the equation for \(y\):
\[ y = -2 \cdot \frac{x}{x+1} + 1 \]
This simplifies to:
\[ y = -\frac{2x}{x+1} + 1 = 1 - \frac{2x}{x+1} \]
Rearranging gives:
\[ y - 1 = -\frac{2x}{x+1} \]
Multiplying through by \((x+1)\) yields:
\[ (x+1)(y-1) + 2x = 0 \]

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