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

Options

A. log 6
B. `log((2)/(3))`
C. `log((3)/(2))`
D. log 8

Answer

To solve the definite integral \( \int_{0}^{\log 2} \frac{e^x}{1 + e^x} \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = 1 + e^x \). Then, we differentiate both sides:
\[ dt = e^x \, dx \quad \Rightarrow \quad dx = \frac{dt}{e^x} \]
From our substitution, we can express \( e^x \) in terms of \( t \):
\[ e^x = t - 1 \]

### Step 2: Change the limits of integration
Next, we need to change the limits of integration according to our substitution:
- When \( x = 0 \):
\[ t = 1 + e^0 = 1 + 1 = 2 \]
- When \( x = \log 2 \):
\[ t = 1 + e^{\log 2} = 1 + 2 = 3 \]

Thus, the limits change from \( x: 0 \) to \( \log 2 \) into \( t: 2 \) to \( 3 \).

### Step 3: Substitute in the integral
Now we can substitute into the integral:
\[ \int_{0}^{\log 2} \frac{e^x}{1 + e^x} \, dx = \int_{2}^{3} \frac{t - 1}{t} \cdot \frac{dt}{e^x} \]
Since \( e^x = t - 1 \), we have:
\[ \int_{2}^{3} \frac{t - 1}{t} \cdot \frac{dt}{t - 1} = \int_{2}^{3} \frac{1}{t} \, dt \]

### Step 4: Evaluate the integral
The integral \( \int \frac{1}{t} \, dt \) is:
\[ \log t \]
Thus, we evaluate:
\[ \int_{2}^{3} \frac{1}{t} \, dt = \left[ \log t \right]_{2}^{3} = \log 3 - \log 2 \]

### Step 5: Simplify the result
Using the properties of logarithms:
\[ \log 3 - \log 2 = \log \left( \frac{3}{2} \right) \]

### Final Answer
Therefore, the value of the integral is:
\[ \int_{0}^{\log 2} \frac{e^x}{1 + e^x} \, dx = \log \left( \frac{3}{2} \right) \]

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