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

Options

A. `log (e^x + sqrt(e^(2x) - 1)) - sec^(-1) e^x`
B. `log (e^x + sqrt(e^(zx) - 1)) + sec^(-1) e^x`
C. `log (e^x - sqrt(e^(zx) - 1)) - sec^(-1) e^x`
D. none

Answer

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

### Step 1: Simplifying the Integral
We start with the integral:
\[ I = \int \sqrt{\frac{e^x - 1}{e^x + 1}} \, dx \]
To simplify the expression under the square root, we can multiply the numerator and denominator by \(-1\):
\[ I = \int \sqrt{\frac{-(e^x - 1)}{-(e^x + 1)}} \, dx = \int \sqrt{\frac{1 - e^x}{1 + e^x}} \, dx \]

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

### Step 3: Separating the Integral
We can separate the integral:
\[ I = \int \frac{\sqrt{1 - t}}{t \sqrt{1 + t}} \, dt \]

### Step 4: Further Simplification
We can rewrite this integral as:
\[ I = \int \frac{\sqrt{1 - t}}{\sqrt{1 + t}} \cdot \frac{1}{t} \, dt \]

### Step 5: Using Trigonometric Substitution
To simplify the integral further, we can use the substitution \( t = \sin^2(\theta) \), which gives us:
\[ dt = 2 \sin(\theta) \cos(\theta) \, d\theta \]
Then, we have:
\[ I = \int \frac{\sqrt{1 - \sin^2(\theta)}}{\sqrt{1 + \sin^2(\theta)}} \cdot \frac{2 \sin(\theta) \cos(\theta)}{\sin^2(\theta)} \, d\theta \]
This simplifies to:
\[ I = 2 \int \frac{\cos^2(\theta)}{\sin^2(\theta) \sqrt{1 + \sin^2(\theta)}} \, d\theta \]

### Step 6: Final Integration
Now we can integrate:
\[ I = 2 \int \cot^2(\theta) \cdot \frac{1}{\sqrt{1 + \sin^2(\theta)}} \, d\theta \]
This integral can be solved using standard integral formulas.

### Step 7: Back Substitution
After finding the integral in terms of \( \theta \), we substitute back to \( t \) and then back to \( x \) using \( t = e^x \).

### Final Result
The final result will be expressed in terms of \( x \).