A curve passes through (2, 0) and the slope of tangents at point P(x, y) equals `((x+1)^(2)+y-3)/((x+1))`.
Find the equation of the curve and area enclosed by the curve and the X-axis in the fourth quadrant.

Answer

To solve the problem, we need to find the equation of the curve given the slope of the tangent at any point \( P(x, y) \) and then calculate the area enclosed by the curve and the X-axis in the fourth quadrant.

### Step 1: Set up the differential equation
The slope of the tangent at point \( P(x, y) \) is given by:
\[ \frac{dy}{dx} = \frac{(x+1)^2 + (y-3)}{(x+1)} \]

### Step 2: Simplify the differential equation
We can rewrite the equation as:
\[ \frac{dy}{dx} = (x + 1) + \frac{y - 3}{(x + 1)} \]
Let \( t = x + 1 \). Then, \( x = t - 1 \) and \( dx = dt \). The equation becomes:
\[ \frac{dy}{dt} = t + \frac{y - 3}{t} \]

### Step 3: Substitute and rearrange
Substituting \( z = y - 3 \), we have:
\[ \frac{dy}{dt} = t + \frac{z}{t} \]
This leads to:
\[ \frac{dz}{dt} + \frac{z}{t} = t \]

### Step 4: Solve the linear differential equation
This is a linear first-order differential equation. The integrating factor \( \mu(t) \) is given by:
\[ \mu(t) = e^{\int \frac{1}{t} dt} = e^{\ln |t|} = t \]
Multiplying through by the integrating factor, we have:
\[ t \frac{dz}{dt} + z = t^2 \]

### Step 5: Integrate both sides
Integrating both sides:
\[ \int \left( t \frac{dz}{dt} + z \right) dt = \int t^2 dt \]
This gives:
\[ z = \frac{t^3}{3} + C \cdot t \]

### Step 6: Substitute back for \( y \)
Recall \( z = y - 3 \) and \( t = x + 1 \):
\[ y - 3 = \frac{(x + 1)^3}{3} + C(x + 1) \]
Thus, we have:
\[ y = \frac{(x + 1)^3}{3} + C(x + 1) + 3 \]

### Step 7: Use the initial condition
The curve passes through the point \( (2, 0) \):
\[ 0 = \frac{(2 + 1)^3}{3} + C(2 + 1) + 3 \]
Calculating:
\[ 0 = \frac{27}{3} + 3C + 3 \implies 0 = 9 + 3C + 3 \implies 3C = -12 \implies C = -4 \]

### Step 8: Final equation of the curve
Substituting \( C \) back into the equation:
\[ y = \frac{(x + 1)^3}{3} - 4(x + 1) + 3 \]
Simplifying:
\[ y = \frac{(x + 1)^3}{3} - 4x - 4 + 3 \]
\[ y = \frac{(x + 1)^3}{3} - 4x - 1 \]

### Step 9: Find the area in the fourth quadrant
The area \( A \) between the curve and the X-axis from \( x = 0 \) to \( x = 2 \):
\[ A = \int_0^2 |y| \, dx \]
Since \( y \) is negative in this interval, we take the negative of the integral:
\[ A = -\int_0^2 \left( \frac{(x + 1)^3}{3} - 4x - 1 \right) dx \]

### Step 10: Evaluate the integral
Calculating the integral:
\[ A = -\left[ \frac{(x + 1)^4}{12} - 2x^2 - x \right]_0^2 \]
Evaluating at the limits:
\[ = -\left( \frac{(3)^4}{12} - 2(2)^2 - 2 \right) + \left( 0 \right) \]
\[ = -\left( \frac{81}{12} - 8 - 2 \right) = -\left( \frac{81}{12} - 10 \right) = -\left( \frac{81 - 120}{12} \right) = \frac{39}{12} = \frac{13}{4} \]

### Final Answer
The area enclosed by the curve and the X-axis in the fourth quadrant is:
\[ \boxed{\frac{13}{4}} \]