If $a+1=0$ or $a=-1$, the given equation becomes a line equation with a solution $x=-2$. Otherwise, if $a+1\ne 0$, let the two roots be $x_1$ and $x_2$, respectively. By the Vieta's theorem, we have $$x_1+x_2=\dfrac{a^2+1}{a+1}=(a-1)+\dfrac{2}{a+1}$$ Because both $x_1$ and $x_2$ are integers, we find $a+1$ must be a divisor of 2. There means $a=1,0,-2,$ or $-3$. - When $a=1$, the two roots are $x=2$ and $x=-1$ - When $a=0$, the two roots are $x=3$ and $x=-2$ - When $a=-2$, no integer solution exists - When $a=-3$, no integer solution exists Therefore the answer is $-1+0+1=\boxed{0}$.