Find the range of f(x)=`(1)/(pi)sin^(-1)x+tan^(-1)+(x+1)/(x^(2)+2x+5)`

Explanation

To find the range of the function \( f(x) = \frac{1}{\pi} \sin^{-1}(x) + \tan^{-1}(x) + \frac{x+1}{x^2 + 2x + 5} \), we will follow a systematic approach:

### Step 1: Determine the Domain
The function consists of three parts: \( \sin^{-1}(x) \), \( \tan^{-1}(x) \), and \( \frac{x+1}{x^2 + 2x + 5} \).

1. **Domain of \( \sin^{-1}(x) \)**: The domain is \( x \in [-1, 1] \).
2. **Domain of \( \tan^{-1}(x) \)**: The domain is \( x \in \mathbb{R} \).
3. **Domain of \( \frac{x+1}{x^2 + 2x + 5} \)**: The denominator \( x^2 + 2x + 5 \) is always positive for all \( x \) (since its discriminant is negative). Thus, its domain is also \( x \in \mathbb{R} \).
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