Determine if $\int_{0}^{1}\sqrt{x}\ln{x} \ dx$ is convergent I'm so utterly lost when it comes to these improper integrals. Everytime i feel like I got the hang of it, I always get the wrong answer.
I can clearly see that the function is not defined for $x=0$. So I need to figure out what happens there by examining the limit
$$\lim_{x\rightarrow0_+}\sqrt{x}\ln{x}.$$
Turns out there does exists a finite value here, which is 0. But how do I show this? Is there any other way to, without even computing the integral, "quickly see" that the function is convergent or divergent?

To get rid of $\sqrt{x}$ you could substitute $u=\sqrt{x}$.
Then $\displaystyle I=\int_0^1 \sqrt{x}\ln(x)\,dx=\int_0^1 u\ln(u^2)2u\,du=4\int_0^1 u^2\ln(u)\,du$
Notice that $u\mapsto u^2\ln(u)$ is continuous on any $[\delta,1]$ with $0<\delta\ll 1$ so the part $\displaystyle \int_{\delta}^1 u^2\ln(u)\,du$ is finite.
Now in the interval $]0,\delta[$ let's study $f(u)=u\ln(u)$.
$f'(u)=\ln(u)+1\to-\infty$ so $f\searrow$ in a neighborhood of $0^+$ namely on $]0,\delta[$ for a small enough $\delta$.
We deduce that $f$ is bounded on this interval: $|u\ln(u)|<C$
[i.e.  in simple words you cannot decrease from $-\infty$, in math language $g(v)=f(\frac 1v)\nearrow$ and upper bounded $g(v)\le 0$ thus converges when $v\to+\infty$, and a convergent continuous function is bounded. ].
So $\displaystyle \bigg|\int_0^\delta u^2\ln(u)\,du\bigg|\le \int_0^\delta u|u\ln(u)|\,du\le \int_0^\delta Cu\,du=\frac 12C\delta^2\ll 1$
