if `int_(log2)^(x) (du)/(e^u-1)^(1/2) =pi/6` then `e^x=`

Options

A. 1
B. 2
C. 4
D. `-1`

Explanation

To solve the given problem, we need to evaluate the definite integral and find the value of \( e^x \). Let's break it down step by step.

### Step 1: Set up the integral
We are given the equation:
\[ \int_{\log 2}^{x} \frac{du}{(e^u - 1)^{1/2}} = \frac{\pi}{6} \]

### Step 2: Substitute to simplify the integral
We will make the substitution:
\[ e^u - 1 = t^2 \]
This implies:
\[ e^u = t^2 + 1 \quad \text{and} \quad du = \frac{2t}{e^u} dt = \frac{2t}{t^2 + 1} dt \]

### Step 3: Change the limits of integration
When \( u = \log 2 \):
\[ e^{\log 2} - 1 = 2 - 1 = 1 \implies t = 1 \]
When \( u = x \):
\[ e^x - 1 = t^2 \implies t = \sqrt{e^x - 1} \]

### Step 4: Rewrite the integral
Now we can rewrite the integral with the new variable:
\[ \int_{1}^{\sqrt{e^x - 1}} \frac{2t}{(t^2 + 1)^{1/2}} dt \]
This simplifies to:
\[ \int_{1}^{\sqrt{e^x - 1}} 2 \frac{t}{(t^2 + 1)^{1/2}} dt \]

### Step 5: Evaluate the integral
The integral \( \int \frac{t}{(t^2 + 1)^{1/2}} dt \) can be evaluated as:
\[ \int 2 \frac{t}{(t^2 + 1)^{1/2}} dt = 2 \sqrt{t^2 + 1} \]
Thus, we have:
\[ 2 \left[ \sqrt{t^2 + 1} \right]_{1}^{\sqrt{e^x - 1}} = 2 \left( \sqrt{e^x} - \sqrt{2} \right) \]

### Step 6: Set the integral equal to \(\frac{\pi}{6}\)
Setting this equal to \(\frac{\pi}{6}\):
\[ 2 \left( \sqrt{e^x} - \sqrt{2} \right) = \frac{\pi}{6} \]

### Step 7: Solve for \(\sqrt{e^x}\)
Dividing both sides by 2:
\[ \sqrt{e^x} - \sqrt{2} = \frac{\pi}{12} \]
Adding \(\sqrt{2}\) to both sides:
\[ \sqrt{e^x} = \sqrt{2} + \frac{\pi}{12} \]

### Step 8: Square both sides to find \(e^x\)
Now squaring both sides:
\[ e^x = \left( \sqrt{2} + \frac{\pi}{12} \right)^2 \]

### Step 9: Final expression for \(e^x\)
Thus, we have:
\[ e^x = 2 + 2 \cdot \sqrt{2} \cdot \frac{\pi}{12} + \left( \frac{\pi}{12} \right)^2 \]