Let $f \in L^1(\mathbb{R})$ and $p$ a polynomial of deg $m$. Does $f / p \in L^1$ imply $x^m f / p \in L^1$ or vice versa?

1. If both $f$ and $f/p$. Are integrable then $h(x)=x^m\frac{f(x)}{p(x)}$ is also integrable. The reason is that $$\lim_{x\rightarrow\pm\infty}\Big|\frac{x^m}{p(x)}\Big|$$ exists, and it is finite. Hence, for some constant $c$ and for all $x$ large enough, say $|x|>A$ for some $A>0$, $$ \Big|\frac{x^m}{p(x)}\Big|\leq c$$ Consequently $$|h|\leq A^m\Big|\frac{f}{p}\Big|\mathbf{1}(|x|\leq A) + c|f|\mathbf{1}(|x|>A)$$ The conclusion then follows by dominated convergence.

1 is false, by example take $f(x) = e^{-x^2}$ and $p(x)=x$. Then $h(x) = f(x) \in L^1$, $\dfrac{f}{p} = \dfrac{e^{-x^2}}{x} \notin L^1.$