Given,

$f\left(x\right)=\frac{x^2+2x+1}{x^2+1}$

$f^{\text{'}}\left(x\right)=\frac{\left(x^2+1\right)(2x+2)-(x^2+2x+1)2x}{{\left(x^2+1\right)}^2}$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{\cancel{2x^3}+2x^2+\cancel{2x}+2-\cancel{2x^3}-4x^2-\cancel{2x}}{{\left(x^2+1\right)}^2}$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{2(1-x^2)}{{\left(x^2+1\right)}^2}=-\frac{2(x+1)(x-1)}{{\left(x^2+1\right)}^2}$

Clearly $f\left(x\right)$ is one-one in $\left(-\infty ,-1\right)$ and also in $(1,\infty )$ but $f\left(x\right)$ is not one-one in $\left(-\infty ,\infty \right)$