`int sqrt(e^(x)-1)dx` is equal to

Options

A. `2[sqrt(e^(x)-1)-tan^(-1) sqrt(e^(x)-1)]+c`
B. `sqrt(e^(x)-1)-tan^(-1) sqrt(e^(x)-1)+c`
C. `sqrt(e^(x)-1)+tan^(-1) sqrt(e^(x)-1)+c`
D. `2[sqrt(e^(x)-1)+tan^(-1) sqrt(e^(x)-1)]+c`

Explanation

To solve the integral \( \int \sqrt{e^x - 1} \, dx \), we will use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Substitution
Let us set:
\[ t^2 = e^x - 1 \]
This implies:
\[ e^x = t^2 + 1 \]

### Step 2: Differentiate
Now, differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(e^x) = \frac{d}{dx}(t^2 + 1) \]
This gives:
\[ e^x = 2t \frac{dt}{dx} \]
From this, we can express \( dx \) in terms of \( dt \):
\[ dx = \frac{e^x}{2t} \, dt = \frac{t^2 + 1}{2t} \, dt \]

### Step 3: Substitute in the Integral
Now substitute \( \sqrt{e^x - 1} \) and \( dx \) into the integral:
\[ \int \sqrt{t^2} \cdot \frac{t^2 + 1}{2t} \, dt = \int t \cdot \frac{t^2 + 1}{2t} \, dt = \int \frac{t^2 + 1}{2} \, dt \]

### Step 4: Simplify the Integral
This simplifies to:
\[ \int \frac{t^2 + 1}{2} \, dt = \frac{1}{2} \int (t^2 + 1) \, dt \]
Now, we can integrate:
\[ \frac{1}{2} \left( \frac{t^3}{3} + t \right) + C = \frac{t^3}{6} + \frac{t}{2} + C \]

### Step 5: Substitute Back
Now, substitute \( t = \sqrt{e^x - 1} \) back into the expression:
\[ \frac{(\sqrt{e^x - 1})^3}{6} + \frac{\sqrt{e^x - 1}}{2} + C \]
This simplifies to:
\[ \frac{(e^x - 1)^{3/2}}{6} + \frac{\sqrt{e^x - 1}}{2} + C \]

### Final Answer
Thus, the integral \( \int \sqrt{e^x - 1} \, dx \) is given by:
\[ \frac{(e^x - 1)^{3/2}}{6} + \frac{\sqrt{e^x - 1}}{2} + C \]