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!
 A: Both integrals diverge.  The first diverges because the integrand behaves as $x^{-5/2}$ as $x \to 0$, which is a non-integrable singularity.  You can see that the second diverges upon substituting $x=e^{u}-1$ - the integrand behaves as $e^{u/2}/u^3$ as $u \to \infty$.
A: A related problem.
1) The integral diverges since as $x\sim 0$
$$\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}\sim \frac{\sqrt{x}}{(1)(x^3) }\sim \frac{1}{x^{5/2}}.$$
Note:
$$ \ln(1+x) = x - \frac{x^2}{2} + \dots. $$
2) For the second integral, just replace $x \leftrightarrow 1/x $, so the integrand will behave as $x\sim 0$ as
$$ \frac{\sqrt{1/x}}{(1+1/x)\ln^3(1+1/x)}= \frac{\sqrt{x}}{(1+x)(\ln^3(1+1/x))}\sim \frac{\sqrt{x}}{(x)(\ln^3(1/x))} = -\frac{1}{\sqrt{x}\ln^3(x)}.$$
A: 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 $\int_1^\infty\frac1{1+x}\,\mathrm{d}x$
