Let  $f:R\to R,f\left(x\right)=x+\text{ln}(1+x^2)$  then

$f\left(x\right)=x+\ln (1+x^2)\Rightarrow f\text{'}\left(x\right)=1+\frac{2x}{1+x^2}=\frac{{(x+1)}^2}{1+x^2}\ge 0\forall x\in R$

So,  $f\left(x\right)$  is injective and at  $x=-1$, tangent is not parallel to any of the chords
Again

  $$\begin{gathered} {\text{lim}}_{x\to -\infty }f\left(x\right)={\text{lim}}_{x\to -\infty }\text{ln}(e^x\cdot (1+x^2))={\text{lim}}_{x\to -\infty }\text{ln}\left(\frac{1+x^2}{e^{-x}}\right) \\ ={\text{lim}}_{x\to -\infty }\text{ln}\left(\frac{2x}{e^{-x}}\right)={\text{lim}}_{x\to -\infty }\text{ln}\left(\frac{2}{e^{-x}}\right)=-\infty \end{gathered}$$
So, f is surjective also.