If `int(dx)/(1+e^x)=x+Klog(1+e^(x))+C` then K is equal to

Options

A. 1
B. `-1`
C. 2
D. `-2`

Answer

To solve the given problem, we need to evaluate the integral \( I = \int \frac{dx}{1 + e^x} \) and compare it with the expression \( x + K \log(1 + e^x) + C \) to find the value of \( K \).

### Step-by-Step Solution:

1. **Start with the integral**:
\[ I = \int \frac{dx}{1 + e^x} \]

2. **Rewrite the integrand**:
We can express \( e^x \) in terms of \( e^{-x} \):
\[ I = \int \frac{e^{-x}}{e^{-x} + 1} \, dx \]

3. **Substitution**:
Let \( t = 1 + e^{-x} \). Then, the derivative \( dt = -e^{-x} \, dx \) or \( dx = -\frac{dt}{e^{-x}} \). Since \( e^{-x} = t - 1 \), we can rewrite \( dx \):
\[ dx = -\frac{dt}{t - 1} \]

4. **Change of limits**:
Substitute \( e^{-x} \) into the integral:
\[ I = \int \frac{-dt}{t} = -\ln |t| + C \]
Substituting back for \( t \):
\[ I = -\ln |1 + e^{-x}| + C \]

5. **Using logarithmic properties**:
We can express the logarithm:
\[ I = -\ln(1 + e^{-x}) + C = \ln\left(\frac{1}{1 + e^{-x}}\right) + C \]

6. **Further simplification**:
We can rewrite \( e^{-x} \) as \( \frac{1}{e^x} \):
\[ I = \ln\left(\frac{e^x}{e^x + 1}\right) + C \]

7. **Expressing in terms of \( K \)**:
Now, we can express this in the form \( x + K \log(1 + e^x) + C \):
\[ I = x - \ln(1 + e^x) + C \]

8. **Comparing with the given expression**:
From the expression \( x + K \log(1 + e^x) + C \), we can see that:
\[ K = -1 \]

### Final Answer:
Thus, the value of \( K \) is:
\[ \boxed{-1} \]