`int_(0)^(log 2)(e^(x))/(1+e^(x))dx=`

Answer

To solve the integral

\[ \int_{0}^{\log 2} \frac{e^{x}}{1 + e^{x}} \, dx, \]

we will use a substitution method.

### Step 1: Substitution
Let

\[ T = 1 + e^{x}. \]

Then, differentiating both sides gives us

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

### Step 2: Express \( e^{x} \) in terms of \( T \)
From our substitution, we have

\[ e^{x} = T - 1. \]

### Step 3: Change the limits of integration
Now we need to change the limits of integration according to our substitution.

- When \( x = 0 \):
\[ T = 1 + e^{0} = 1 + 1 = 2. \]

- When \( x = \log 2 \):
\[ T = 1 + e^{\log 2} = 1 + 2 = 3. \]

So the new limits of integration are from \( T = 2 \) to \( T = 3 \).

### Step 4: Substitute into the integral
Now we substitute \( e^{x} \) and \( dx \) into the integral:

\[ \int_{0}^{\log 2} \frac{e^{x}}{1 + e^{x}} \, dx = \int_{2}^{3} \frac{T - 1}{T} \cdot \frac{dT}{T - 1} = \int_{2}^{3} \frac{1}{T} \, dT. \]

### Step 5: Evaluate the integral
Now we can evaluate the integral:

\[ \int \frac{1}{T} \, dT = \log T. \]

Thus, we have:

\[ \int_{2}^{3} \frac{1}{T} \, dT = \log T \Big|_{2}^{3} = \log 3 - \log 2. \]

### Step 6: Simplify the result
Using the logarithmic property \( \log a - \log b = \log \frac{a}{b} \):

\[ \log 3 - \log 2 = \log \frac{3}{2}. \]

### Final Answer
Therefore, the value of the integral is

\[ \int_{0}^{\log 2} \frac{e^{x}}{1 + e^{x}} \, dx = \log \frac{3}{2}. \]

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