`int(dx)/(sqrt(1-e^(2x)))=?`

Options

A. `log |e^(x)+sqrt(e^(2x)-1)|+C`
B. `log |e^(-x)+sqrt(e^(-2x)-1)|+C`
C. `-log |e^(-x)+sqrt(e^(-2x)-1)|+C`
D. None of these

Explanation

To solve the integral \( \int \frac{dx}{\sqrt{1 - e^{2x}}} \), we will proceed step by step.

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

### Step 2: Substitute \( e^{-x} \)
To simplify the integral, we can use the substitution \( t = e^{-x} \). Then, we have:
\[ dx = -\frac{dt}{t} \]
This substitution transforms the integral:
\[ I = \int \frac{-\frac{dt}{t}}{\sqrt{1 - (e^{-x})^2}} = \int \frac{-dt}{t \sqrt{1 - t^2}} \]

### Step 3: Simplify the Integral
Now, we can rewrite the integral as:
\[ I = -\int \frac{dt}{t \sqrt{1 - t^2}} \]

### Step 4: Use a Known Integral Formula
We know from integral tables that:
\[ \int \frac{dt}{t \sqrt{1 - t^2}} = \ln |t + \sqrt{t^2 - 1}| + C \]
Thus, we can write:
\[ I = -\ln |t + \sqrt{t^2 - 1}| + C \]

### Step 5: Substitute Back for \( t \)
Now we substitute back \( t = e^{-x} \):
\[ I = -\ln |e^{-x} + \sqrt{(e^{-x})^2 - 1}| + C \]

### Step 6: Simplify the Expression
We simplify the expression:
\[ I = -\ln \left| e^{-x} + \sqrt{e^{-2x} - 1} \right| + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{dx}{\sqrt{1 - e^{2x}}} = -\ln \left| e^{-x} + \sqrt{e^{-2x} - 1} \right| + C \]

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