The question is: exists a natural number $n \geq 2$such that
$$ \displaystyle\int_{0}^{+ \infty} \displaystyle\frac{\ln r}{(1 + r^2)^{n}} r dr< \infty ?$$
I am trying to do this : i know that exists $A>0$ such that $ln r < r , \forall r > A$
note that $\displaystyle\int_{0}^{+ \infty} \displaystyle\frac{\ln r}{(1 + r^2)^{n}} r dr = \displaystyle\int_{0}^{A} \displaystyle\frac{\ln r}{(1 + r^2)^{n}} r dr + \displaystyle\int_{A}^{+ \infty} \displaystyle\frac{\ln r}{(1 + r^2)^{n}} r dr$
The second integral converges for all $n \geq 2$ because for $r > A $ we have $\displaystyle\frac{\ln r}{(1 + r^2)^{n}} r < \displaystyle\frac{r^2}{(1+r^2)^n} < \displaystyle\frac{r^2}{r^4} = \displaystyle\frac{1}{r^2}$ .
and the integral $\displaystyle\int_{A}^{+ \infty} \displaystyle\frac{1}{r^2} dr$ converges. The problem is the first integral..
Someone can give me a hint? Any help is welcome .
Thanks in advance