jmc

Since $q(x)$ is a quadratic, and we have a horizontal asymptote at $y=0,$ we know that $p(x)$ must be linear.

Since we have a hole at $x=1,$ there must be a factor of $x-1$ in both $p(x)$ and $q(x).$ Additionally, since there is a vertical asymptote at $x=-1,$ the denominator $q(x)$ must have a factor of $x+1.$ Then, $p(x)=a(x-1)$ and $q(x)=b(x+1)(x-1),$ for some constants $a$ and $b.$

Since $p(2)=1$, we have $a(2-1)=1$ and hence $a=1$. Since $q(2)=3$, we have $b(2+1)(2-1)=3$ and hence $b=1$.

So $p(x)=x-1$ and $q(x)=(x+1)(x-1)=x^2-1,$ so $p(x)+q(x)=\boxed{x^2+x-2}.$