For which values of parameter $c\in\mathbb{R}$ is $$\int_0^{\infty}\left(\frac{x}{\ln x}\right)^c\frac{1}{1+x^{4c}}\ dx < \infty$$ Using the inequality $\ln x > \frac{x}{1+x}$ it can be shown that $$\int_0^{a}\left(\frac{x}{\ln x}\right)^c\frac{1}{1+x^{4c}}\ dx < \int_0^{a}\left(\frac{x}{\frac{x}{1+x}}\right)^c\frac{1}{1+x^{4c}}\ dx = \int_0^{a}\frac{(1+x)^c}{1+x^{4c}}\ dx < \infty \ \ \forall c\in\mathbb{R}$$
But I don't know what next, how to nicely check convergence of $$\int_a^{\infty}\left(\frac{x}{\ln x}\right)^c\frac{1}{1+x^{4c}}\ dx$$ is respect to $c$. Got any tips?
Around $+\infty$, the integrand behaves like $\frac{x^{c- 4c}}{\ln^{c} x}$ then we need $-3c<-1$ (the P-test) meaning $c>1/3$.
The above condition is necessary, it remains to prove that if $c>1/3$ the integral converges for $a>1$ which is easy.
If $a<1$, we must have $c<1$ due to the singularity around zero.
Conclusion : the integral mentioned in the title converge iff $c\in (1/3,1)$.