`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 can follow these steps:

### Step 1: Substitution
Let \( e^x - 1 = t^2 \). Then, we have:
\[ e^x = t^2 + 1 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(e^x) = \frac{d}{dx}(t^2 + 1) \implies e^x \, dx = 2t \, dt \]
Thus, we can express \( dx \) as:
\[ dx = \frac{2t}{e^x} \, dt = \frac{2t}{t^2 + 1} \, dt \]

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

### Step 3: Simplify the Integral
We can split the integrand:
\[ \frac{2t^2}{t^2 + 1} = 2 - \frac{2}{t^2 + 1} \]
Thus, the integral becomes:
\[ \int \frac{2t^2}{t^2 + 1} \, dt = \int \left(2 - \frac{2}{t^2 + 1}\right) dt \]

### Step 4: Integrate
Now we can integrate term by term:
\[ \int 2 \, dt - 2 \int \frac{1}{t^2 + 1} \, dt = 2t - 2 \tan^{-1}(t) + C \]

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

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

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