Test for convergence of improper integrals $\int_{0}^{1}\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}dx$ and $\int_{1}^{\infty}\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}dx$ I need to test if, integrals below, either converge or diverge:
1) $\displaystyle\int_{0}^{1}\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}dx$
2) $\displaystyle\int_{1}^{\infty}\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}dx$
I tried comparing with $\displaystyle\int_{0}^{1}\frac{1}{(1+x)\ln^3(1+x)}dx$, $\displaystyle\int_{0}^{1}\frac{\sqrt{x}}{(1+x)}dx$ but ended up with nothing. 
Do you have any suggestions? Thanks!

Near $0$, $\log(1+x)=x(1+O(x))$ so
$$
\frac{\sqrt{x}}{(1+x)\log^3(1+x)}=x^{-5/2}(1+O(x))
$$
and because $-5/2\le-1$, the integral in 1) does not converge.

$$
\left(\frac{\sqrt{x}}{\log^3(1+x)}\right)^{1/3}=\frac{x^{1/6}}{\log(1+x)}
$$
By L'Hospital,
$$
\begin{align}
\lim_{x\to\infty}\frac{x^{1/6}}{\log(1+x)}
&=\lim_{x\to\infty}\frac{\frac16x^{-5/6}}{1/(1+x)}\\
&=\lim_{x\to\infty}\frac16\left(x^{-5/6}+x^{1/6}\right)\\[12pt]
&=\infty
\end{align}
$$
Therefore,
$$
\lim_{x\to\infty}\frac{\sqrt{x}}{\log^3(1+x)}=\infty
$$
Thus, the integral in 2) does not converge by comparison to $\int_1^\infty\frac1{1+x}\,\mathrm{d}x$
