convergence of $\large \int_1^{+\infty} \frac{\ln(x)}{\sqrt{1+x}}dx$ I'm trying to determine if the underneath integral converges or not.
$\large \int_1^{+\infty}  \frac{\ln(x)}{\sqrt{1+x}}dx$

I know that in the interval [1, $+\infty$)  the enqualities:

$ \frac{\ln(x)}{\sqrt{1+x}} \leq  \frac{\ln(x)}{\sqrt{x}}$  holds.

so if  $\int_1^{+\infty} \frac{\ln(x)}{\sqrt{x}} dx$  converges then   so does   $\large \int_1^{+\infty}  \frac{\ln(x)}{\sqrt{1+x}}dx$

Does my anlysis hold?

$$I=\int_{a}^{\infty} \frac{dx}{x^{\beta}}<\infty$$ if $\beta>1$ diverges if $\beta<1$, because in your case $\beta=1/2<1$.
Hence the given integral will diverge like $$\int_{1}^{\infty} \frac{dx}{x^{0.99}}$$
