jmc

Let $$y=\frac{x^2+x+1}{x^2+1}.$$Then $x^2+x+1=y(x^2+1),$ which we write as $$(y-1)x^2-x+(y-1)=0.$$If $y=1,$ then this simplifies to $x=0.$ In other words, $p(0)=1.$ Otherwise, the equation above is a quadratic, whose discriminant is $$1-4(y-1{)}^2=-4y^2+8y-3.$$For a given value of $y,$ the quadratic has a real solution in $x$ if and only if this discriminant is nonnegative. Thus, we want to solve the inequality $$-4y^2+8y-3\ge 0.$$We can factor this as $$-(2y-3)(2y-1)\ge 0.$$The solution to this inequality is $\frac{1}{2}\le y\le \frac{3}{2}.$ Note that this interval includes the value of $p(0)=1$ that we found above, so the range of the function is $\boxed{\left[\frac{1}{2},\frac{3}{2}\right]}.$