We have the improper integral $$\int_0^\infty \frac{\tanh(x)}{\left(1+x^2\right)^a}\, \mathrm{d}x$$ and we need to find all the values of $a$, such that the above integral converges.
We know that $0< \tanh(x) < 1, \forall x\in (0,\infty)$. Also, for every $a\ge 1$ we know that $$\frac{1}{\left(1+x^2\right)^a} \le \frac{1}{1+x^2}.$$
- For $a\ge 1$ by comparison test we have: $$0\le \int_0^\infty \frac{\tanh(x)}{\left(1+x^2\right)^a }\, \mathrm{d}x\le \int_{0}^\infty \frac{1}{\left(1+x^2\right)^a}\,\mathrm{d}x \le \int_0^\infty \frac{1}{1+x^2}\,\mathrm{d}x.$$ The last integral converges, so our initial integral converges, as well.
I have some difficulty with the cases $0<a<1$ and and $a \le 0$. I tried to use the comparison test with some known functions, even the Bernoulli's inequality without any success. It seems I miss something fundamental.
Hint. The integrand is a continuous function over $[0,\infty)$. There is no problem near $0$. A potential issue is as $x \to \infty$, but in this case we have $$ \frac{\tanh(x)}{\left(1+x^2\right)^a } \sim \frac1{x^{2a} } $$ giving a convergence for $a>\frac12$ and a divergence for $a<\frac12$.