$\int_{0}^{\infty} \frac{\ln(x^2+1)}{x^\alpha} dx$
Now I am a bit confused here, I know that for a very small integer $(x)$, $\ln(x^2+1)$ acts very similar to $x^2$, so I can solve it for $(0,1)$
but checking from $1$ to infinity, I know that $\ln(x^2+1)$ acts like $x^\epsilon$
but I dont really know how to go from there, Ive tried convergence tests but didnt got a defenite limit. treid using $f(x)=x^\epsilon / x^\alpha $ but the limit of the quotient is 0.
the answer is $1 < \alpha < 3$
if anyone knows how to solve it :) thank you