If `int_(ln2)^(x) (e^(x) - 1)^(-1) dx = ln'3/2` then what is the value of x ?

Options

A. `e^(2)`
B. `1/e`
C. `ln 4`
D. `1`

Explanation

To solve the problem, we need to evaluate the integral and find the value of \( x \) such that:

\[ \int_{\ln 2}^{x} \frac{1}{e^t - 1} \, dt = \frac{\ln 3}{2} \]

### Step 1: Rewrite the Integral
The integral can be rewritten as:

\[ \int_{\ln 2}^{x} (e^t - 1)^{-1} \, dt = \int_{\ln 2}^{x} \frac{1}{e^t - 1} \, dt \]

### Step 2: Change of Variables
Let \( u = e^t - 1 \). Then, differentiating gives:

\[ du = e^t \, dt \quad \Rightarrow \quad dt = \frac{du}{e^t} = \frac{du}{u + 1} \]

### Step 3: Change the Limits of Integration
When \( t = \ln 2 \):

\[ u = e^{\ln 2} - 1 = 2 - 1 = 1 \]

When \( t = x \):

\[ u = e^x - 1 \]

Thus, the integral becomes:

\[ \int_{1}^{e^x - 1} \frac{1}{u} \cdot \frac{1}{u + 1} \, du \]

### Step 4: Partial Fraction Decomposition
We can express \( \frac{1}{u(u + 1)} \) as:

\[ \frac{1}{u(u + 1)} = \frac{A}{u} + \frac{B}{u + 1} \]

Multiplying through by \( u(u + 1) \) gives:

\[ 1 = A(u + 1) + Bu \]

Setting \( u = 0 \):

\[ 1 = A(0 + 1) \quad \Rightarrow \quad A = 1 \]

Setting \( u = -1 \):

\[ 1 = B(-1) \quad \Rightarrow \quad B = -1 \]

Thus,

\[ \frac{1}{u(u + 1)} = \frac{1}{u} - \frac{1}{u + 1} \]

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

\[ \int_{1}^{e^x - 1} \left( \frac{1}{u} - \frac{1}{u + 1} \right) du = \left[ \ln |u| - \ln |u + 1| \right]_{1}^{e^x - 1} \]

This simplifies to:

\[ \left[ \ln(e^x - 1) - \ln(e^x) \right] - \left[ \ln(1) - \ln(2) \right] \]

### Step 6: Substitute Limits
Evaluating the limits gives:

\[ \ln(e^x - 1) - x - (0 - \ln 2) = \ln(e^x - 1) - x + \ln 2 \]

### Step 7: Set Equal to \( \frac{\ln 3}{2} \)
Set the expression equal to \( \frac{\ln 3}{2} \):

\[ \ln(e^x - 1) - x + \ln 2 = \frac{\ln 3}{2} \]

### Step 8: Rearranging
Rearranging gives:

\[ \ln(e^x - 1) = x - \ln 2 + \frac{\ln 3}{2} \]

### Step 9: Exponentiate Both Sides
Exponentiating both sides yields:

\[ e^x - 1 = e^{x - \ln 2 + \frac{\ln 3}{2}} \]

### Step 10: Solve for \( x \)
This leads to:

\[ e^x - 1 = \frac{3}{2} e^{x - \ln 2} = \frac{3}{2} \cdot \frac{e^x}{2} \]

From here, we can simplify and solve for \( x \):

\[ e^x - 1 = \frac{3}{4} e^x \]

This simplifies to:

\[ \frac{1}{4} e^x = 1 \quad \Rightarrow \quad e^x = 4 \quad \Rightarrow \quad x = \ln 4 \]

### Final Answer
Thus, the value of \( x \) is:

\[ \boxed{\ln 4} \]