Convergence of the following integral as $\alpha\in\mathbb{R}$
$$\int_{0}^{+\infty}\left(x^2+\log\left(1+\frac{1}{x^2}\right)\right)^{\alpha}-x^{2\alpha}\,dx$$
I splitted it into: $$\int_{0}^{2}f(x)\,dx \,\,+\,\,\int_{2}^{+\infty}f(x)\,dx$$
Now for $x\to+\infty$ we have, after some calculations, that $f(x) = \mathcal{O}\left(\frac{1}{x^{2(2-\alpha)}}\right)$, which converges when $\alpha <\frac{3}{2}$
How to interpret $f(x)$ in a neighborhood of $0$?
Suppose $0<\alpha<2$. Then near 0, one has $$\left ( x^2+\log(1+1/x^2) \right )^\alpha \leq \left ( x^2+\log(1+1/x^2) \right )^2 \leq C \left ( \log(1+1/x^2) \right )^2.$$ So it is sufficient to show $\int (\log(1+1/x^2))^2 dx$ is convergent near 0. By integration by parts, one has $$\int (\log(1+1/x^2))^2 dx=x (\log (1+1/x^2))^2+4\int \dfrac{\log(1+1/x^2)dx}{x^2+1},$$ which is convergent near 0 if $\int \log(1+1/x^2)dx$ is (by comparison). Another integration by parts gives $$\int \log(1+1/x^2)dx=x \log(1+1/x^2)+\int \dfrac{2dx}{x^2+1},$$ which is convergent near 0.
If $\alpha<0$, then $x^{2\alpha}$ is integrable near 0 if and only if $\alpha>-1/2$. So the improper integral is finite for all $-1/2<\alpha<3/2$.