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

Options

A. `x-log[1+sqrt(1-e^(2x))]+c`
B. `x+log[1+sqrt(1-e^(2x))]+c`
C. `log[1+sqrt(1-e^(2x))]-x+c`
D. none of these

Answer

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

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

### Step 2: Substitute \( e^x \)
Let \( t = e^x \). Then, \( dt = e^x \, dx \) or \( dx = \frac{dt}{t} \). The integral becomes:
\[ I = \int \frac{1}{\sqrt{1 - t^2}} \cdot \frac{dt}{t} \]

### Step 3: Rewrite the Integral
This can be rewritten as:
\[ I = \int \frac{1}{t \sqrt{1 - t^2}} \, dt \]

### Step 4: Use Trigonometric Substitution
We can use the substitution \( t = \sin(\theta) \), which gives \( dt = \cos(\theta) \, d\theta \). The integral becomes:
\[ I = \int \frac{1}{\sin(\theta) \sqrt{1 - \sin^2(\theta)}} \cos(\theta) \, d\theta \]
Since \( \sqrt{1 - \sin^2(\theta)} = \cos(\theta) \), we have:
\[ I = \int \frac{1}{\sin(\theta)} \, d\theta \]

### Step 5: Integrate
The integral of \( \frac{1}{\sin(\theta)} \) is:
\[ I = \ln |\tan(\frac{\theta}{2})| + C \]

### Step 6: Back Substitute
Now we need to back substitute \( \theta \) in terms of \( t \) and \( x \):
Since \( t = \sin(\theta) \), we have \( \theta = \arcsin(t) = \arcsin(e^x) \). Thus:
\[ I = \ln |\tan(\frac{1}{2} \arcsin(e^x))| + C \]

### Step 7: Final Expression
We can express this in terms of \( e^x \) and simplify further if necessary, but this is the general form of the solution.

### Summary of the Solution
The integral evaluates to:
\[ I = \ln |\tan(\frac{1}{2} \arcsin(e^x))| + C \]