If `int(x(x-1))/((x^(2)+1)(x+1)sqrt(x^(3)+x^(2)+x))dx`
`=(1)/(2)log_(e)|(sqrt(f(x))-1)/(sqrt(f(x))+1)|-tan^(-1)sqrt(f(x))+C,` then
The value of `lim_(x to oo) tan^(-1)sqrt(f(x))` is

Options

A. `pi//2`
B. `pi//4`
C. `pi`
D. `2pi`

Explanation

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to \infty} \tan^{-1} \sqrt{f(x)} \]

where \( f(x) = \frac{x + 1}{x + 1} \) as derived from the given integral.

### Step-by-Step Solution:

1. **Identify \( f(x) \)**:
From the video transcript, we see that \( f(x) \) is derived from the integral and is given as:
\[ f(x) = \frac{x + 1}{x + 1} \]
Simplifying this, we find:
\[ f(x) = 1 \]

2. **Calculate \( \sqrt{f(x)} \)**:
Now, we need to find \( \sqrt{f(x)} \):
\[ \sqrt{f(x)} = \sqrt{1} = 1 \]

3. **Evaluate \( \tan^{-1} \sqrt{f(x)} \)**:
Now we substitute \( \sqrt{f(x)} \) into the inverse tangent function:
\[ \tan^{-1} \sqrt{f(x)} = \tan^{-1}(1) \]

4. **Find the limit**:
We need to find the limit as \( x \) approaches infinity:
\[ \lim_{x \to \infty} \tan^{-1} \sqrt{f(x)} = \tan^{-1}(1) \]
The value of \( \tan^{-1}(1) \) is:
\[ \tan^{-1}(1) = \frac{\pi}{4} \]

5. **Final Result**:
Thus, the final result is:
\[ \lim_{x \to \infty} \tan^{-1} \sqrt{f(x)} = \frac{\pi}{4} \]

### Conclusion:
The value of \( \lim_{x \to \infty} \tan^{-1} \sqrt{f(x)} \) is \( \frac{\pi}{4} \).