The equation of the curve for which the slope of the tangent at any point is given by `(x+y+1)((dy)/(dx))=1` is (where, c is an arbitrary constant)

Options

A. `xy=e^(x)-c`
B. `xy =ce^(y)+2`
C. `x=ce^(y)-y-2`
D. `x=e^(y)+y-c`

Answer

To solve the problem, we need to find the equation of the curve for which the slope of the tangent at any point is given by the equation:

\[ (x+y+1) \frac{dy}{dx} = 1 \]

### Step 1: Rearranging the Equation

First, we can rearrange the given equation to express \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{1}{x+y+1} \]

**Hint:** To isolate \(\frac{dy}{dx}\), divide both sides of the equation by \(x+y+1\).

### Step 2: Separating Variables

Next, we can separate the variables \(y\) and \(x\):

\[ dy = \frac{1}{x+y+1} dx \]

**Hint:** Remember that separating variables allows us to integrate both sides independently.

### Step 3: Integrating Both Sides

Now, we will integrate both sides. However, to integrate the right-hand side, we need to express it in a more manageable form. We can rewrite it as:

\[ dy = \frac{1}{x+y+1} dx \]

This is a bit tricky because we have \(y\) in the denominator. To make it easier, we can use substitution. Let \(u = x + y + 1\), then \(du = dx + dy\) or \(dy = du - dx\).

Now, substituting \(dy\) in terms of \(du\):

\[ du - dx = \frac{1}{u} dx \]

Rearranging gives:

\[ du = \left(1 + \frac{1}{u}\right) dx \]

This leads to:

\[ du = \frac{u + 1}{u} dx \]

Now, we can integrate both sides:

\[ \int du = \int \frac{1}{u} dx \]

**Hint:** Use the substitution method to simplify the integration process.

### Step 4: Solving the Integrals

Integrating both sides, we get:

\[ u = \ln|u| + C \]

Substituting back \(u = x + y + 1\):

\[ x + y + 1 = \ln|x + y + 1| + C \]

**Hint:** Remember to include the constant of integration \(C\) after integrating.

### Step 5: Rearranging the Equation

Now, we can rearrange this equation to express it in a more standard form. We can express it as:

\[ x + y + 1 - \ln|x + y + 1| = C \]

**Hint:** This step involves rearranging the equation to isolate the terms.

### Final Result

Thus, the equation of the curve is:

\[ x + y + 1 - \ln|x + y + 1| = C \]