Evaluate : `int (1)/(sqrt(1-e^(2x))) " dx "`

Options

A. `log |e^(-x) +sqrt(e^(-x)-1)|+c`
B. `-log |e^(-x) +sqrt(e^(-2x)+1)|+c`
C. `log |e^(-x) +sqrt(e^(-2x)-1)|+c`
D. `-log |e^(-x) +sqrt(e^(-2x)-1)|+c`

Explanation

To evaluate the integral

\[ I = \int \frac{1}{\sqrt{1 - e^{2x}}} \, dx, \]

we will follow these steps:

### Step 1: Rewrite the Integral
We start by factoring out \( e^{2x} \) from the square root in the denominator:

\[ I = \int \frac{1}{\sqrt{1 - e^{2x}}} \, dx = \int \frac{1}{\sqrt{e^{2x}(e^{-2x} - 1)}} \, dx. \]

### Step 2: Simplify the Integral
This can be simplified further:

\[ I = \int \frac{1}{e^x \sqrt{e^{-2x} - 1}} \, dx = \int \frac{e^{-x}}{\sqrt{e^{-2x} - 1}} \, dx. \]

### Step 3: Substitution
Let \( t = e^{-x} \). Then, \( dt = -e^{-x} \, dx \) or \( dx = -\frac{dt}{t} \). Substituting these into the integral gives:

\[ I = \int \frac{t}{\sqrt{t^2 - 1}} \left(-\frac{dt}{t}\right) = -\int \frac{1}{\sqrt{t^2 - 1}} \, dt. \]

### Step 4: Integrate
The integral \( \int \frac{1}{\sqrt{t^2 - 1}} \, dt \) is a standard integral, which results in:

\[ -\int \frac{1}{\sqrt{t^2 - 1}} \, dt = -\ln |t + \sqrt{t^2 - 1}| + C. \]

### Step 5: Substitute Back
Now, substituting back \( t = e^{-x} \):

\[ I = -\ln |e^{-x} + \sqrt{(e^{-x})^2 - 1}| + C. \]

### Step 6: Simplify the Expression
This can be simplified further:

\[ I = -\ln \left( e^{-x} + \sqrt{e^{-2x} - 1} \right) + C. \]

### Final Result
Thus, the final result of the integral is:

\[ I = -\ln \left( e^{-x} + \sqrt{e^{-2x} - 1} \right) + C. \]