Evaluate `int_(0)^(1)(e^(-x)dx)/(1+e^(x))`

Explanation

To evaluate the integral

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

we will follow a systematic approach.

### Step 1: Rewrite the Integral

We start with the integral:

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

Notice that we can rewrite \( e^{-x} \) as \( \frac{1}{e^{x}} \):

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

### Step 2: Substitution

Next, we perform the substitution \( z = e^{x} \). Then, we have:

\[ dz = e^{x} \, dx \quad \Rightarrow \quad dx = \frac{dz}{z}. \]

Now we need to change the limits of integration. When \( x = 0 \), \( z = e^{0} = 1 \), and when \( x = 1 \), \( z = e^{1} = e \). Thus, the limits change from \( x: 0 \to 1 \) to \( z: 1 \to e \).

Substituting these into the integral gives:

\[ I = \int_{1}^{e} \frac{1}{z(1 + z)} \cdot \frac{dz}{z} = \int_{1}^{e} \frac{1}{z^2(1 + z)} \, dz. \]

### Step 3: Simplifying the Integrand

Now we simplify the integrand:

\[ \frac{1}{z^2(1 + z)} = \frac{1}{z^2} - \frac{1}{1 + z}. \]

Thus, we can split the integral:

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

### Step 4: Evaluate Each Integral

Now we evaluate the two integrals separately:

1. For \( \int \frac{1}{z^2} \, dz \):

\[ \int \frac{1}{z^2} \, dz = -\frac{1}{z}. \]

2. For \( \int \frac{1}{1 + z} \, dz \):

\[ \int \frac{1}{1 + z} \, dz = \ln(1 + z). \]

### Step 5: Combine the Results

Putting it all together, we have:

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

Now we evaluate at the limits:

At \( z = e \):

\[ -\frac{1}{e} - \ln(1 + e), \]

At \( z = 1 \):

\[ -\frac{1}{1} - \ln(1 + 1) = -1 - \ln(2). \]

Thus, we have:

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

### Step 6: Final Calculation

This simplifies to:

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

### Final Answer

Thus, the final result of the integral is:

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