Existence or evaluation of$\int^{+\infty}_{0}\frac{\log(x)}{1 + x\log^{2}(x)}dx$. Determine if the following integral exists and is finite:
$$\int^{+\infty}_{0}\dfrac{\log(x)}{1 + x\log^{2}(x)}dx$$
So, usually to determine if an improper integral exists and is finite you would have to determine if the limit exists and is finite:
$$\lim_{b\to +\infty}\int^{b}_{a}\dfrac{\log(x)}{1 + x\log^{2}(x)}dx$$
But that would require me to calculate the primitive of the function, and it doesn't look appealing.
When reading the calculus book made by my professor, I found a theorem that stated that for improper integrals of the type $[a, b[$ if $f(x)\ge 0$ in that interval, then the integral existed.
Now: $\dfrac{\log(x)}{1 + x\log^{2}(x)} \ge 0 $ only when $x \ge 1$, which is only a subset of the interval. Also in zero we have a divergence for log(x).
So my question is: does the theorem I mentioned above apply in this case, and if it does, why does it apply (considering the problems I've mentioned).
If it doesn't apply, what's the best course of action to find the primitive of the function?
Thanks for the help.
 A: $$\int_{0}^{+\infty}\frac{\log x}{1+x\log^2 x}\,dx = \int_{-\infty}^{+\infty}\frac{t}{e^{-t}+ t^2}\,dt $$
is blatantly divergent since for any $M>0$
$$ \int_{0}^{M}\frac{t}{e^{-t}+t^2}\,dt\geq \int_{0}^{M}\frac{t}{1+t^2}\,dt = \frac{1}{2}\log(M^2+1).$$
A: For $x\in (0,1)$ we have $\log x\le 0$  and  $1\le 1+x\log^2(x)$ we get 
$$\log(x) \le \frac{\log(x)}{1+x\log^2(x)}\le 0$$
Then this yields, 
$$-1 =[x\log x-x]_0^1= \int^{1}_{0}\log(x)dx\le\color{red}{\int^{1}_{0}\dfrac{\log(x)}{1 + x\log^{2}(x)}dx}\le 0$$
We have, $$\color{blue}{\lim_{x\to\infty}\frac{x\log x\log x}{1+x\log^2( x)}=1 }$$ 
Therefore there exists $c>1$ such that for every $x>c$ we have, 
$$\color{red}{\frac{1}{2}\dfrac{1}{ x\log(x)}\le \lim_{x\to\infty}\frac{x\log x\log x}{1+x\log^2( x)}\le \frac{3}{2}\dfrac{1}{ x\log(x)}}  $$
Then, use the Bertrand Criteria to  see that 
$$\int^{+\infty}_{c}\dfrac{\log(x)}{1 + x\log^{2}(x)}dx\ge \frac{1}{2}\int^{+\infty}_{c}\dfrac{1}{ x\log^{}(x)}dx =\infty$$
diverges.
Now For $x\in(1,c)$ it converges as the integrand is continuous on $[1,c].$
Then your integral diverges since
$$\int^{+\infty}_{0}\dfrac{\log(x)}{1 + x\log^{2}(x)}dx =\int^{+\infty}_{c}\dfrac{\log(x)}{1 + x\log^{2}(x)}dx+\int^{c}_{1}\dfrac{\log(x)}{1 + x\log^{2}(x)}dx+\int^{1}_{0}\dfrac{\log(x)}{1 + x\log^{2}(x)}dx =\infty$$
