If `I=inte^(-x)log(e^x+1)dx ,t h e nIe q u a l` `a+(e^(-x)+1)log(e^x+1)+C` `a+(e^x+1)log(e^x+1)+C` `a-(e^(-x)+1)log(e^x+1)+C` none of these

Options

A. `x+(e^(-x)+1)log(e^(x)+1)+C`
B. `x+(e^(x)+1)log(e^(x)+1)+C`
C. `x-(e^(-x)+1)log(e^(x)+1)+C`
D. none of these

Explanation

To solve the integral \( I = \int e^{-x} \log(e^x + 1) \, dx \), we will use integration by parts. Let's denote:

- \( u = \log(e^x + 1) \) (which we will differentiate)
- \( dv = e^{-x} \, dx \) (which we will integrate)

### Step 1: Differentiate \( u \) and Integrate \( dv \)

1. Differentiate \( u \):
\[ du = \frac{1}{e^x + 1} \cdot e^x \, dx = \frac{e^x}{e^x + 1} \, dx \]

2. Integrate \( dv \):
\[ v = \int e^{-x} \, dx = -e^{-x} \]

### Step 2: Apply Integration by Parts

Using the integration by parts formula:
\[ I = uv - \int v \, du \]
we substitute \( u \), \( du \), \( v \), and \( dv \):
\[ I = \log(e^x + 1)(-e^{-x}) - \int (-e^{-x}) \left(\frac{e^x}{e^x + 1}\right) \, dx \]
This simplifies to:
\[ I = -\log(e^x + 1)e^{-x} + \int \frac{1}{e^x + 1} \, dx \]

### Step 3: Solve the Remaining Integral

Now we need to solve:
\[ \int \frac{1}{e^x + 1} \, dx \]
We can use the substitution \( t = e^x + 1 \), which gives \( dt = e^x \, dx \) or \( dx = \frac{dt}{e^x} = \frac{dt}{t - 1} \). Thus, we have:
\[ \int \frac{1}{t} \cdot \frac{dt}{t - 1} \]
This integral can be solved using partial fractions:
\[ \frac{1}{t(t - 1)} = \frac{1}{t} + \frac{1}{1 - t} \]
Integrating gives:
\[ \int \frac{1}{t} \, dt - \int \frac{1}{1 - t} \, dt = \log|t| - \log|1 - t| + C \]
Substituting back \( t = e^x + 1 \):
\[ \int \frac{1}{e^x + 1} \, dx = \log(e^x + 1) - \log(-e^x) + C \]

### Step 4: Combine the Results

Now we combine everything:
\[ I = -\log(e^x + 1)e^{-x} + \log(e^x + 1) - \log(-e^x) + C \]

### Final Answer

Thus, the final result can be expressed as:
\[ I = -\log(e^x + 1)e^{-x} + \log(e^x + 1) + C \]