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

Options

A. `2sec^(-1)sqrt(2e^(x))+c`
B. `-2"tan"^(-1)(1)/(sqrt(2e^(x)-1))+c`
C. `2sec^(-1)(sqrt(2)e^(x))+c`
D. `2tan^(-1)sqrt(2e^(x)-1)+c`

Explanation

To evaluate the integral

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

we will use a substitution method. Here are the steps:

### Step 1: Substitution
Let
\[ t^2 = 2e^x - 1. \]
Then, differentiating both sides with respect to \(x\), we get:
\[ 2e^x \, dx = 2t \, dt. \]
This simplifies to:
\[ e^x \, dx = t \, dt. \]

### Step 2: Express \(dx\) in terms of \(dt\)
From the previous equation, we can express \(dx\) as:
\[ dx = \frac{t}{e^x} \, dt. \]
Now, we need to express \(e^x\) in terms of \(t\). From our substitution \(t^2 = 2e^x - 1\), we can rearrange it to find \(e^x\):
\[ 2e^x = t^2 + 1 \quad \Rightarrow \quad e^x = \frac{t^2 + 1}{2}. \]

### Step 3: Substitute back into the integral
Now substituting \(dx\) and \(e^x\) into the integral \(I\):
\[ I = \int \frac{dx}{\sqrt{t^2}} = \int \frac{\frac{t}{\frac{t^2 + 1}{2}} \, dt}{t} = \int \frac{2}{t^2 + 1} \, dt. \]

### Step 4: Evaluate the integral
The integral \(\int \frac{2}{t^2 + 1} \, dt\) is a standard integral:
\[ I = 2 \tan^{-1}(t) + C, \]
where \(C\) is the constant of integration.

### Step 5: Substitute back for \(t\)
Recall that \(t = \sqrt{2e^x - 1}\). Therefore, we substitute back:
\[ I = 2 \tan^{-1}(\sqrt{2e^x - 1}) + C. \]

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