<s>[INST] Find the range of the given function $f$: $f(x) = \frac{1}{\pi} \left(\sin^{-1} x + \tan^{-1} x\right) + \frac{x+1}{x^2 + 2x + 5} $ 
Find the range of the following function. 
  $$\cfrac{1}{\pi} \left(\sin^{-1} x + \tan^{-1} x\right) + \cfrac{x+1}{x^2 + 2x + 5} $$
  where $\sin^{-1}x \ $ and $ \ \tan^{-1}x $ are inverse trigonometric functions.

This is what I've tried yet.
Let : $\cfrac{1}{\pi} \left(\sin^{-1} x + \tan^{-1} x\right) = g(x)$ and $ \cfrac{x+1}{x^2 + 2x + 5} = h(x)$ .
Therefore, 
$$f(x) = g(x) + h(x)$$ 
I can write $h(x)$ as : $$h(x) = \cfrac{x+1}{(x+1)^2 + 4} = \cfrac{1}{(x+1) + \cfrac{4}{x+1}}$$
So, the function $f(x)$ becomes: 
$$f(x) = \cfrac{1}{\pi} \left(\sin^{-1} x +\tan^{-1} x\right) + \cfrac{1}{(x+1) + \cfrac{4}{x+1}}$$
Domain of $g(x)$ is $[-1,1] \cap \mathbb{R} = [-1,1]$ 
Domain of $h(x)$ is $\mathbb{R} - \{-1\}$ 
Therefore, $dom(f(x)) = (-1,1]$ 
Now, finding maximum value of $h(x)$ 
$$\begin{align} 
\cfrac{dh(x)}{dx} =& \cfrac{d}{dx} \left( (x+1) + \cfrac{4}{x+1} \right) \\
= & 1 - \cfrac{4}{(x+1)^2}\end{align}$$
Putting $h'(x)$ equal to zero to find critical points, I got: 
$$\begin{align} 1 - \cfrac{4}{(x+1)^2} =& 0 \\
(x+1) =& \pm 2  \\
\boxed{x =  1 \ \text{OR} \ x = -3} \rightarrow \textsf{CRITICAL POINTS} \end{align}$$
How to go further from here? Is this method the perfect one for solving this type of questions? Or is there any other trick to go with it?
 [/INST] Note that the domain of $g(x)=\frac{1}{\pi}(\sin^{-1}x+\tan^{-1}x)$ is $[-1,1]$ and the domain of $h(x)=\frac{x+1}{x^2+2x+5}$ is $(-\infty,\infty)$. Thus the domain of $f(x)=g(x)+h(x)$ is $[-1,1]$. Note that also 
$$ g'(x)=\frac{1}{\pi}\left(\frac{1}{\sqrt{1-x^2}}+\frac{1}{x^2+1}\right)>0$$
and
$$ h'(x)=\frac{4-(x+1)^2}{(x+1)^2+4}>0 \text{ for }x\in[-1,1]$$
and hence $f(x)=g(x)+h(x)$ is increasing in $[-1,1]$. So the range of $f(x)$ is
$$[f(-1),f(1)]=[-\frac{3}{4},1].$$
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