The value of `int_(0) ^(1) (dx )/( e ^(x) + e)` is equal to

Options

A. `1/e log ((1 + e)/(2))`
B. ` log ((1+e )/(2 ))`
C. `1/e log (1 +e)`
D. `log ((2)/( 1 + e))`

Answer

To solve the integral

\[ I = \int_{0}^{1} \frac{dx}{e^x + e}, \]

we will follow these steps:

### Step 1: Substitute \( t = e^x + e \)

Let \( t = e^x + e \). Then, differentiating both sides gives:

\[ dt = e^x \, dx \implies dx = \frac{dt}{e^x}. \]

From our substitution, we can express \( e^x \) in terms of \( t \):

\[ e^x = t - e. \]

### Step 2: Change the limits of integration

When \( x = 0 \):

\[ t = e^0 + e = 1 + e. \]

When \( x = 1 \):

\[ t = e^1 + e = e + e = 2e. \]

Thus, the limits change from \( x = 0 \) to \( x = 1 \) into \( t = 1 + e \) to \( t = 2e \).

### Step 3: Rewrite the integral

Substituting \( dx \) and changing the limits, we have:

\[ I = \int_{1 + e}^{2e} \frac{1}{t} \cdot \frac{dt}{t - e}. \]

### Step 4: Simplify the integral

We can rewrite the integrand as:

\[ \frac{1}{t(t - e)} = \frac{1}{e} \left( \frac{1}{t - e} - \frac{1}{t} \right). \]

Thus, we can express the integral as:

\[ I = \frac{1}{e} \int_{1 + e}^{2e} \left( \frac{1}{t - e} - \frac{1}{t} \right) dt. \]

### Step 5: Integrate

Now we can integrate term by term:

\[ I = \frac{1}{e} \left[ \ln |t - e| - \ln |t| \right]_{1 + e}^{2e}. \]

### Step 6: Evaluate the limits

Evaluating at the limits \( t = 2e \) and \( t = 1 + e \):

\[ I = \frac{1}{e} \left[ \ln |2e - e| - \ln |2e| - \left( \ln |1 + e - e| - \ln |1 + e| \right) \right]. \]

This simplifies to:

\[ I = \frac{1}{e} \left[ \ln |e| - \ln |2e| - \left( \ln |1| - \ln |1 + e| \right) \right]. \]

### Step 7: Simplify further

This can be simplified to:

\[ I = \frac{1}{e} \left[ 1 - \ln 2 - \left( 0 - \ln(1 + e) \right) \right]. \]

Thus, we have:

\[ I = \frac{1}{e} \left[ 1 - \ln 2 + \ln(1 + e) \right]. \]

### Final Answer

Therefore, the value of the integral is:

\[ I = \frac{1}{e} \left[ \ln(1 + e) - \ln 2 \right] = \frac{1}{e} \ln \left( \frac{1 + e}{2} \right). \]