find $f(x) \in \Bbb Z[x]$ s.t $f(x)-\frac {p(x)}{x^k} \in P=\{\frac {(x^2+1)q(x)}{x^k}| k\geq 0, q(x) \in \Bbb Z[x]\}$ where $p(x)\in \Bbb Z[x]$.

Question

Given $\displaystyle\frac {p(x)}{x^k}$, find $f(x) \in \Bbb Z[x]$ s.t. $f(x)-\displaystyle\frac {p(x)}{x^k} \in P=\{\frac {(x^2+1)q(x)}{x^k}\mid k\geq 0, q(x) \in \Bbb Z[x]\}$ where $p(x)\in \Bbb Z[x]$.

what I have found that $f(x)=\displaystyle\frac {(x^2+1)q(x)+p(x)}{x^k}$ but there is no guarantee that $x^2+1\mid p(x)$ so how do I get $f(x)\in \Bbb Z[x]$?