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?

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
$$[f(-1),f(1)]=[-\frac{3}{4},1].$$
