$(1+x^2)T = 1$ in $\mathscr D'(\mathbb{R})$

Question

I only know that the solutions of $xT=0$ in $D'(\mathbb{R})$ are in the form $c\delta_0$, but I can't figure out how to find general solution to $(1+x^2)T = 1$.
Any ideas?

Answer 1

Hint: if $f\in C^\infty$ and $fT=0$ in the sense of distributions, then $supp T\subset f^{-1}(\{0\})$, i.e. support of $T$ is a subset of the set of zeroes of $f$.

Second hint: $\frac{1}{1+x^2}\in C^{\infty}(\Bbb R)$.