What's wrong with my solution to the integral, we need to find for which values of $a$ the integral converges
$$
\int_{0}^{\infty}\frac{\arctan\left(\sqrt{6x}\right)}{x^{\left(1-a\right)}\left(\ln\left(3+x^{4}\right)\right)^{2}}dx
$$
I separated the calculation in
$$
\int_{0}^{5}\frac{\arctan\left(\sqrt{6x}\right)}{x^{\left(1-a\right)}\left(\ln\left(3+x^{4}\right)\right)^{2}}dx\ +\ \int_{5}^{\infty}\frac{\arctan\left(\sqrt{6x}\right)}{x^{\left(1-a\right)}\left(\ln\left(3+x^{4}\right)\right)^{2}}dx
$$
The first integral is asymptotically equal to
$$
\frac{\arctan\left(\sqrt{6x}\right)}{x^{\left(1-a\right)}\left(\ln\left(3+x^{4}\right)\right)^{2}}\sim_{x\to0^+}\frac{\sqrt{6x}}{x^{1-a}(x^4+2)^2}\sim_{x\to0^+} \frac{\sqrt{6x}}{x^{1-a}(1+2(x^4+1))}=\frac{\sqrt{6x}}{x^{1-a}(2x^4+3)}
$$
And at this point i get stuck because i can't get to a form like $\frac{1}{x^k}$
The second integral should be:
$$
\frac{\arctan\left(\sqrt{6x}\right)}{x^{\left(1-a\right)}\left(\ln\left(3+x^{4}\right)\right)^{2}}\sim_{x\to\infty}\frac{\frac \pi2}{x^{1-a}(\ln(x^4))^2}=\frac\pi{32}\cdot\frac1{x^{1-a}\ln^2 x}
$$
Which converges for $a\leq0$
I am still really not sure of anything i did, so i'm sorry if there are dumb steps