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

Options

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

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{e^x}{\sqrt{e^{2x} - 1}} \, dx \]
We can rewrite \( e^{2x} \) as \( (e^x)^2 \). Thus, we have:
\[ \int \frac{e^x}{\sqrt{(e^x)^2 - 1}} \, dx \]

### Step 2: Substitution
Let \( t = e^x \). Then, the differential \( dt = e^x \, dx \) or \( dx = \frac{dt}{t} \). Substituting these into the integral gives:
\[ \int \frac{t}{\sqrt{t^2 - 1}} \cdot \frac{dt}{t} = \int \frac{1}{\sqrt{t^2 - 1}} \, dt \]

### Step 3: Recognize the Standard Integral
The integral \( \int \frac{1}{\sqrt{t^2 - 1}} \, dt \) is a standard integral, which evaluates to:
\[ \ln |t + \sqrt{t^2 - 1}| + C \]

### Step 4: Substitute Back
Now, we substitute back \( t = e^x \):
\[ \ln |e^x + \sqrt{(e^x)^2 - 1}| + C \]
This simplifies to:
\[ \ln (e^x + \sqrt{e^{2x} - 1}) + C \]

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

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