II. Find all integer values of $a$ for which the equation $(a+1) x^{2}-\left(a^{2}+1\right) x$ $+2 a^{3}-6=0$ has integer roots.
When $a=-1$, the original equation has integer roots.
When $a \neq-1$, $\triangle=-7 a^{4}-8 a^{3}+2 a^{2}+24 a+25$. We have
$$
\left\{\begin{array}{c}
\triangle \leqslant-84 a+24 a+25<0, \text { when } a \geqslant 2 \\
\triangle \leqslant-10|a|^{2}-24|a|+25<0, \\
\text { when } a \leqslant-2
\end{array}\right.
$$

Therefore, for the known quadratic equation to have integer roots, $a$ must satisfy $-1 \leqslant a \leqslant 1$. When $a=1$, $\triangle=36$, the equation has integer roots; when $a=0$, $\triangle=25$, the equation has integer roots. In summary, the integer values of $a$ are $-1,0,1$.