What is `int(dx)/(2^(x)-1)` equal to ?

Options

A. `ln(2^(x)-1)+C`
B. `(ln(1-2^(-x)))/(ln2)+C`
C. `(ln(2^(-x)-1))/(2ln2)+C`
D. `(ln(1+2^(-x)))/(ln2)+C`

Answer

To solve the integral \( \int \frac{dx}{2^x - 1} \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{2^x - 1} \]

### Step 2: Factor out \( 2^x \)
We can rewrite the denominator:
\[ 2^x - 1 = 2^x \left(1 - 2^{-x}\right) \]
Thus, we can express the integral as:
\[ I = \int \frac{dx}{2^x \left(1 - 2^{-x}\right)} \]

### Step 3: Rewrite the Integral with the Common Factor
Now we can rewrite the integral:
\[ I = \int \frac{2^{-x} \, dx}{1 - 2^{-x}} \]

### Step 4: Use Substitution
Let \( t = 1 - 2^{-x} \). Then, we differentiate \( t \):
\[ dt = \frac{d}{dx}(1 - 2^{-x}) = 2^{-x} \ln(2) \, dx \]
From this, we can express \( dx \):
\[ dx = \frac{dt}{2^{-x} \ln(2)} \]

### Step 5: Substitute Back into the Integral
Now we substitute \( dx \) back into the integral:
\[ I = \int \frac{2^{-x} \cdot \frac{dt}{2^{-x} \ln(2)}}{t} = \int \frac{dt}{\ln(2) t} \]

### Step 6: Integrate
The integral simplifies to:
\[ I = \frac{1}{\ln(2)} \int \frac{dt}{t} \]
The integral of \( \frac{1}{t} \) is:
\[ \int \frac{dt}{t} = \ln |t| + C \]
Thus, we have:
\[ I = \frac{1}{\ln(2)} \ln |t| + C \]

### Step 7: Substitute Back for \( t \)
We substitute back \( t = 1 - 2^{-x} \):
\[ I = \frac{1}{\ln(2)} \ln |1 - 2^{-x}| + C \]

### Final Answer
The final result for the integral is:
\[ \int \frac{dx}{2^x - 1} = \frac{\ln |1 - 2^{-x}|}{\ln(2)} + C \]