The solution of `(e^(x)+1)y dy +(y+1)dx=0` is `e^(x+y) =k(y+1) (1+e^(x))`.

Answer

To solve the differential equation \((e^{x}+1)y \, dy + (y+1) \, dx = 0\), we will follow the steps of separation of variables and integration.

### Step 1: Rearranging the Equation
We start with the given equation:
\[ (e^{x}+1)y \, dy + (y+1) \, dx = 0 \]
Rearranging it gives:
\[ (e^{x}+1)y \, dy = -(y+1) \, dx \]
Dividing both sides by \((y+1)(e^{x}+1)\):
\[ \frac{y \, dy}{y+1} = -\frac{dx}{e^{x}+1} \]

### Step 2: Separating Variables
Now we have separated the variables \(y\) and \(x\):
\[ \frac{y \, dy}{y+1} = -\frac{dx}{e^{x}+1} \]

### Step 3: Integrating Both Sides
Next, we integrate both sides:
\[ \int \frac{y \, dy}{y+1} = -\int \frac{dx}{e^{x}+1} \]

#### Left Side Integration:
For the left side:
\[ \int \frac{y \, dy}{y+1} = \int \left(1 - \frac{1}{y+1}\right) dy = y - \ln|y+1| + C_1 \]

#### Right Side Integration:
For the right side:
\[ -\int \frac{dx}{e^{x}+1} = -\ln|e^{x}+1| + C_2 \]

### Step 4: Combining Results
Combining the results from both integrations, we get:
\[ y - \ln|y+1| = -\ln|e^{x}+1| + C \]
where \(C = C_2 - C_1\).

### Step 5: Exponentiating Both Sides
To eliminate the logarithm, we exponentiate both sides:
\[ e^{y - \ln|y+1|} = e^{- \ln|e^{x}+1| + C} \]
This simplifies to:
\[ \frac{e^{y}}{y+1} = k \cdot \frac{1}{e^{x}+1} \]
where \(k = e^{C}\).

### Step 6: Rearranging the Equation
Rearranging gives:
\[ e^{y} = k(y+1)(e^{x}+1) \]

### Step 7: Final Form
Finally, we can express this as:
\[ e^{x+y} = k(y+1)(1+e^{x}) \]

### Conclusion
Thus, we have shown that the solution of the differential equation \((e^{x}+1)y \, dy + (y+1) \, dx = 0\) is indeed:
\[ e^{x+y} = k(y+1)(1+e^{x}) \]