Đáp án:
$\begin{array}{l}
\Delta > 0\\
\Rightarrow {\left( {m + 1} \right)^2} + 4\left( {{m^2} + 1} \right) > 0\\
\Rightarrow {m^2} + 2m + 1 + 4{m^2} + 4 > 0\\
\Rightarrow 5{m^2} + 2m + 5 > 0\left( {đúng\,\forall m} \right)\\
Theo\,viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m + 1\\
{x_1}{x_2} = - {m^2} - 1
\end{array} \right.\\
\Rightarrow P = \frac{{ - {m^2} - 1}}{{m + 1}} = - \frac{{{m^2} - 1 + 2}}{{m + 1}}\left( {m \ne - 1} \right)\\
= - \left( {m - 1} \right) + \frac{2}{{m + 1}}\\
P \in Z\\
\Rightarrow 2 \vdots \left( {m + 1} \right)\\
\Rightarrow \left( {m + 1} \right) \in Ư\left( 2 \right) = {\rm{\{ }} - 2; - 1;1;2\} \\
\Rightarrow m \in {\rm{\{ }} - 3; - 2;0;1\} \left( {tmdk} \right)
\end{array}$
