Any idea for sketching $y=g(x)$ where $\tan (g(x))=\frac {x}{1+x^2}$, $g(0)=\pi$ and $-\infty\lt x\lt\infty$?

Question

My attempt. I used the identity $\arctan(\alpha)\pm\arctan(\beta)=\arctan\left(\frac{\alpha\pm \beta}{1\mp \alpha\beta}\right)$ to rewrite $\arctan\left(\frac{x}{1+x^2}\right)$ as a sum of two $\arctan$'s, but I found that it did not help much with the sketch. Then I have been stuck until now. Any hint would be appreciated. Thank you!

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Remark.
In case that someone wants to sketch some similar but more challenging function, the last part of the original question is to sketch $h(x)$ where $\tan(h(x))=\frac {x}{1-x^2}$, $-\infty\lt x\lt\infty$, and $h(0)=\pi$.
Last, this problem has been solved, but any new approach is always welcome. Thanks!

Answer 1

$$\frac{d(\arctan(\frac{x}{1+x^2}))}{dx}=\frac{1}{1+\frac{x^2}{(1+x^2)^2}}.\frac{1+x^2-2x^2}{(1+x^2)^2}$$ $$f'(x)=\frac{1-x^2}{x^4+3x^2+1}$$ $$f'(x)=0,x=\pm1$$ Also $f(0)=0$

and $\lim_{x\to\pm\infty}{f(x)}=0$

Therefore the graph must be one of these:

But we know that $f(x)>0$ when $x>0$ and $f(x)<0$ when $x<0$ So it is the first one.

Answer 2

$g(x)=\tan^{-1} \frac{x}{1+x^2}$, $g(0)=0$, $g(x)$ is an odd function. $g(\pm \infty)=0$ $$g'(x)=\frac{1-x^2}{x^4+3x^2+1}=0\implies x=\pm 1 $$ $g'(x)$ changes sign from negative to positive when we pass $x=-1$ so there is minimu and similarly max at $x=1$. Hence $f_{max}=f(1)=\tan^{-1}(1/2)=0.4634, f_{min}=-0.4634.$ Therefore $g(x)$ a curve which is continuous and odd. It passes through origin vanishes asymptotically and has a max and min at $x=\pm 1.$ See the sketch of $g(x)$ in the fig. below $g(x)$