Extending a rational function by continuity

Idea. A function defined on $\mathbb{R}\setminus\{a\}$ can be extended by continuity at $a$ exactly when it has a finite limit at $a$. The extension is then defined by that limit.

Computation. For $x\neq-1$,
$$f(x)=\frac{1+x}{(x+1)(x^2-x+1)}=\frac1{x^2-x+1}.$$
The denominator $x^2-x+1$ has no real root (its discriminant is $-3<0$), and at $x=-1$ it equals $1+1+1=3$. Hence $\lim_{x\to-1}f(x)=\dfrac13$.

Conclusion. Setting $f(-1)=\frac13$ gives a continuous function on $\mathbb{R}$, namely $x\mapsto\frac1{x^2-x+1}$.