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?
 A: For quickly finding the limit; apply L'Hospital's rule to $$\lim_{x \to 0} \frac{\ln(x)}{\frac{1}{\sqrt x}}$$
To get $$\lim_{x \to 0} \frac{\ln(x)}{\frac{1}{\sqrt x}} =\lim_{x \to 0} \frac{\frac 1x}{\frac{-1}{2 x \sqrt x}}=\lim_{x \to 0} \left(-2\sqrt x \right)=0$$
A: 
I thought it would be instructive to present a way forward that relies on elementary, pre-calculus tools only.  To that end we proceed.


To show that $\sqrt x\log(x) \to 0$ as $x\to 0$, we simply exploit the inequality
$$\frac{x-1}{x}\le\log(x)\le x-1 \tag1$$ 
which I showed using only the limit definition of the exponential function and Bernoulli's inequality in THIS ANSWER.  Note for $0<x\le 1$, we have from $(1)$
$$\frac{x-1}{x}\le \log(x)\le 0 \tag2$$
Replacing $x$ with $x^\alpha$ in $(2)$, multiplying by $\sqrt x$, and using $\log(x^\alpha)=\alpha \log(x)$, we find that for $0<x\le 1$ and $\alpha >0$
$$\frac{\sqrt x-x^{1/2-\alpha}}\alpha\le \sqrt x\log(x)\le 0$$
Restricting $0<\alpha<1/2$ (e.g., Take $\alpha=1/4$.), applying the squeeze theorem yields coveted limit 
$$\lim{x\to 0}\sqrt x\log(x)=0$$
as was to be shown!

Note that by enforcing the substitution $x\to x^2$ and using integration by parts, we find that 
$$\begin{align}
\lim_{\epsilon\to 0}\int_{\epsilon}^1 \sqrt{x}\log(x)\,dx&=\lim_{\epsilon\to 0}\left.\left(\frac29 x^{3/2}(3\log(x)-2)\right)\right|_{\epsilon}^1\\\\
&=\lim_{\epsilon\to 0}\left(-\frac49 -\frac29 \epsilon^{3/2}(3\log(\epsilon)-2)\right)\\\\
&=-\frac49
\end{align}$$

So, the integral converges and the value to which it converges is $-\frac49$.

A: 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 $\displaystyle \bigg|\int_0^\delta u^2\ln(u)\,du\bigg|\le \int_0^\delta u|u\ln(u)|\,du\le \int_0^\delta Cu\,du=\frac 12C\delta^2\ll 1$
