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 use a substitution method. Let's proceed step by step.

### Step 1: Substitution
Let \( e^x = z \). Then, we differentiate both sides to find \( dx \):
\[ dx = \frac{dz}{z}. \]
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 \( x = 1 \) into \( z = 1 \) to \( z = e \).

### Step 2: Rewrite the Integral
Now, substitute \( e^x \) and \( dx \) into the integral:
\[ I = \int_{1}^{e} \frac{e^{-x}}{1 + e^{x}} \cdot \frac{dz}{z}. \]
Since \( e^{-x} = \frac{1}{z} \) and \( 1 + e^x = 1 + z \), we can rewrite the integral as:
\[ I = \int_{1}^{e} \frac{\frac{1}{z}}{1 + z} \cdot \frac{dz}{z} = \int_{1}^{e} \frac{1}{z(1 + z)} \, dz. \]

### Step 3: Simplify the Integral
Now we can simplify the integrand:
\[ I = \int_{1}^{e} \frac{1}{z(1 + z)} \, dz. \]
We can use partial fraction decomposition:
\[ \frac{1}{z(1 + z)} = \frac{A}{z} + \frac{B}{1 + z}. \]
Multiplying through by \( z(1 + z) \) gives:
\[ 1 = A(1 + z) + Bz. \]
Setting \( z = 0 \) gives \( A = 1 \). Setting \( z = -1 \) gives \( B = -1 \). Thus, we have:
\[ \frac{1}{z(1 + z)} = \frac{1}{z} - \frac{1}{1 + z}. \]

### Step 4: Integrate
Now we can integrate:
\[ I = \int_{1}^{e} \left( \frac{1}{z} - \frac{1}{1 + z} \right) dz. \]
This separates into two integrals:
\[ I = \int_{1}^{e} \frac{1}{z} \, dz - \int_{1}^{e} \frac{1}{1 + z} \, dz. \]
Calculating these integrals:
\[ \int \frac{1}{z} \, dz = \ln |z| \quad \text{and} \quad \int \frac{1}{1 + z} \, dz = \ln |1 + z|. \]
Thus,
\[ I = \left[ \ln z \right]_{1}^{e} - \left[ \ln(1 + z) \right]_{1}^{e}. \]

### Step 5: Evaluate the Limits
Now we evaluate the limits:
\[ I = \left( \ln e - \ln 1 \right) - \left( \ln(1 + e) - \ln(1 + 1) \right). \]
This simplifies to:
\[ I = (1 - 0) - \left( \ln(1 + e) - \ln 2 \right). \]
Thus,
\[ I = 1 - \ln(1 + e) + \ln 2. \]

### Final Result
Therefore, the value of the integral is:
\[ I = 1 + \ln 2 - \ln(1 + e). \]