If `int(dx)/(sqrt(e^(x)-1))=2tan^(-1)(f(x))+C`, (where `x gt0` and C is the constant of integration ) then the range of `f(x)` is

Options

A. `(0, oo)`
B. `[0, oo)`
C. `[1, oo)`
D. `(1, oo)`

Answer

To solve the given problem, we need to analyze the integral equation provided and find the range of the function \( f(x) \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ \int \frac{dx}{\sqrt{e^x - 1}} = 2 \tan^{-1}(f(x)) + C \]

2. **Substitute \( e^x - 1 \) with a new variable**:
Let \( t^2 = e^x - 1 \). Therefore, we have:
\[ e^x = t^2 + 1 \]

3. **Differentiate to find \( dx \)**:
Differentiating \( e^x \) gives:
\[ \frac{d}{dx}(e^x) = e^x \implies dx = \frac{2t}{e^x} dt \]
Substituting \( e^x = t^2 + 1 \):
\[ dx = \frac{2t}{t^2 + 1} dt \]

4. **Substitute \( dx \) in the integral**:
Now substitute \( dx \) into the integral:
\[ \int \frac{dx}{\sqrt{e^x - 1}} = \int \frac{2t}{(t^2 + 1) \sqrt{t^2}} dt = \int \frac{2}{t^2 + 1} dt \]

5. **Integrate**:
The integral \( \int \frac{2}{t^2 + 1} dt \) is known:
\[ = 2 \tan^{-1}(t) + C \]

6. **Back-substitute \( t \)**:
Recall that \( t = \sqrt{e^x - 1} \):
\[ 2 \tan^{-1}(\sqrt{e^x - 1}) + C \]

7. **Set the two expressions equal**:
From the original equation:
\[ 2 \tan^{-1}(f(x)) + C = 2 \tan^{-1}(\sqrt{e^x - 1}) + C \]
This implies:
\[ f(x) = \sqrt{e^x - 1} \]

8. **Determine the range of \( f(x) \)**:
- As \( x \) approaches \( 0 \):
\[ e^0 - 1 = 0 \implies \sqrt{0} = 0 \]
- As \( x \) approaches \( \infty \):
\[ e^x - 1 \to \infty \implies \sqrt{e^x - 1} \to \infty \]
Therefore, the range of \( f(x) \) is:
\[ [0, \infty) \]

### Final Answer:
The range of \( f(x) \) is \( [0, \infty) \).