If `I=int(tan^(-1)(e^(x)))/(e^(x)+e^(-x))dx=([tan^(-1)(f(x))]^(2))/(2)+C` (where C is the constant of integration), then the range of `y=f(x) AA x in R` is

Options

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

Answer

To solve the problem step by step, we need to analyze the integral and find the function \( f(x) \) from the given equation.

### Step 1: Write down the integral
We start with the integral given in the question:
\[ I = \int \frac{\tan^{-1}(e^x)}{e^x + e^{-x}} \, dx \]

### Step 2: Simplify the denominator
The denominator can be rewritten:
\[ e^x + e^{-x} = \frac{e^{2x} + 1}{e^x} \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{\tan^{-1}(e^x)}{\frac{e^{2x} + 1}{e^x}} \, dx = \int \frac{e^x \tan^{-1}(e^x)}{e^{2x} + 1} \, dx \]

### Step 3: Use substitution
Let \( t = \tan^{-1}(e^x) \). Then, we differentiate:
\[ \frac{dt}{dx} = \frac{1}{1 + (e^x)^2} \cdot e^x = \frac{e^x}{1 + e^{2x}} \]
This implies:
\[ e^x \, dx = (1 + e^{2x}) \, dt \]

### Step 4: Substitute in the integral
Now, substituting \( e^x \, dx \) in the integral:
\[ I = \int t \, dt \]
The integral of \( t \) is:
\[ I = \frac{t^2}{2} + C \]

### Step 5: Substitute back for \( t \)
Now, substituting back \( t = \tan^{-1}(e^x) \):
\[ I = \frac{(\tan^{-1}(e^x))^2}{2} + C \]

### Step 6: Equate with the given expression
We are given that:
\[ I = \frac{(\tan^{-1}(f(x)))^2}{2} + C \]
From this, we can deduce:
\[ \tan^{-1}(f(x)) = \tan^{-1}(e^x) \]

### Step 7: Solve for \( f(x) \)
Taking the tangent of both sides gives:
\[ f(x) = e^x \]

### Step 8: Determine the range of \( f(x) \)
The function \( f(x) = e^x \) is an exponential function. The range of \( e^x \) as \( x \) varies over all real numbers \( \mathbb{R} \) is:
\[ (0, \infty) \]

### Final Answer
Thus, the range of \( y = f(x) \) for \( x \in \mathbb{R} \) is:
\[ (0, \infty) \]