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

Options

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

Explanation

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:
\[ I = \int \frac{dx}{2^x - 1} \]
We can rewrite the denominator:
\[ I = \int \frac{dx}{2^x - 1} = \int \frac{dx}{2^x (1 - \frac{1}{2^x})} \]

### Step 2: Make a Substitution
Let \( t = 1 - 2^{-x} \). Then, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = \frac{d}{dx}(1 - 2^{-x}) = 0 + 2^{-x} \ln(2) = \frac{\ln(2)}{2^x} \]
Thus, we can express \( dx \) in terms of \( dt \):
\[ dx = \frac{2^x}{\ln(2)} dt \]

### Step 3: Substitute in the Integral
Now substitute \( dx \) and \( 2^x \) in terms of \( t \):
\[ 2^x = \frac{1}{1 - t} \quad \text{(since \( t = 1 - 2^{-x} \))} \]
Substituting these into the integral gives:
\[ I = \int \frac{\frac{2^x}{\ln(2)} dt}{\frac{1}{1 - t} - 1} = \int \frac{\frac{1}{(1 - t) \ln(2)}}{t} dt \]

### Step 4: Simplify the Integral
This simplifies to:
\[ I = \frac{1}{\ln(2)} \int \frac{dt}{t(1 - t)} \]

### Step 5: Partial Fraction Decomposition
We can use partial fractions:
\[ \frac{1}{t(1 - t)} = \frac{1}{t} + \frac{1}{1 - t} \]
Thus, we have:
\[ I = \frac{1}{\ln(2)} \left( \int \frac{1}{t} dt + \int \frac{1}{1 - t} dt \right) \]

### Step 6: Integrate
Integrating gives:
\[ I = \frac{1}{\ln(2)} \left( \ln |t| - \ln |1 - t| + C \right) \]
Combining the logarithms:
\[ I = \frac{1}{\ln(2)} \ln \left| \frac{t}{1 - t} \right| + C \]

### Step 7: Substitute Back
Substituting back \( t = 1 - 2^{-x} \):
\[ I = \frac{1}{\ln(2)} \ln \left| \frac{1 - 2^{-x}}{2^{-x}} \right| + C \]
This simplifies to:
\[ I = \frac{1}{\ln(2)} \ln \left| \frac{2^x - 1}{2^x} \right| + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{dx}{2^x - 1} = \frac{1}{\ln(2)} \ln \left| \frac{2^x - 1}{2^x} \right| + C \]