convergence of $\int_1^{\infty}\frac{lnx}{\sqrt[3]x(x+1)}$ I need help checking if the integral:
$$\int_1^{\infty}\frac{\ln x}{\sqrt[3]x(x+1)}$$
converge or diverge.
I tried comparing to bigger integrals using $\ln x \lt x$ or making the denominator smaller but without success.
any suggestions? 
 A: It is a Bertrand integral.
We have $$\lim_{x\to+\infty}x^\frac 65\frac {\ln (x)}{x^{\frac 13}(1+x)}=0$$
because $\frac 65 <\frac 43$.
$\frac 65>1\implies $ the integral converges.
For $x $ large enough 
$$0 <\frac {\ln (x)}{x^\frac 13 (1+x)}<\frac {1}{x^\frac 65} $$
A: The substitution $x\mapsto z^3$ leads to an absolutely convergent integral:
$$ I = \int_{1}^{+\infty}\frac{\log x}{\sqrt[3]{x}(1+x)}\,dx \stackrel{x\mapsto z^3}{=} \color{blue}{9\int_{1}^{+\infty}\frac{z \log z}{1+z^3}\,dz}\stackrel{z\mapsto t^{-1}}{=}9\int_{0}^{1}\frac{-\log(t)}{1+t^3}\,dt\tag{1}$$
that can be easily computed: since $\int_{0}^{1}t^k(-\log t)\,dt = \frac{1}{(k+1)^2}$,
$$ I = 9\sum_{k\geq 0}\frac{(-1)^k}{(3k+1)^2}\approx\color{blue}{8.56365942}.\tag{2}$$
A: Hint: Compare it to $$\frac{\ln x}{x^{4/3}}.$$
This, incidentally, can be compared to some multiple of$$\frac{\ln(x^{1/1000})}{x^{4/3}},$$which can then be compared to$$\frac1{x^{4/3-1/1000}};$$note that the exponent in the denominator is strictly greater than $1$.
