Why can $\ln x$ be integrated over $[0,1]$, yet $\frac{1}{x}$ can't? Why can $\ln x$ be integrated over $[0,1]$, yet $\frac{1}{x}$ can't?
Both have singularities at $0$, yet we can still integrate $\ln x$ in the specified domain. Why is this?
 A: I assume that you want to compute $$\int_0^1 \ln x d x.$$
This is an improper integral and you compute it (if it exists) as a limit
$$\int_0^1 \ln x d x = \lim_{a\rightarrow 0}\int_a^1 \ln x d x =
\lim_{a\rightarrow 0}[x \ln x - x]_a^1=
\lim_{a\rightarrow 0}(-1-a\ln a +a ) = -1$$
where I have used that $x \ln x - x$ is a antiderivative of $\ln x$ and the well-known limit $\lim_{a\rightarrow 0}(a\ln a) = 0.$ It you want to integrate $1/x$ the same procedure would lead to $\lim_{a\rightarrow 0}(\ln a),\;$ but this does not exist.
A: Whether or not a function with such a singularity can be integrated on an interval or not, is a consequence of how we define the so called improper integral on that interval. I suppose you know how improper integrals are defined, so perhaps you're looking for some intuition.
You can compare this with integrals on intervals stretching to infinity: the integral of $\tfrac{1}{x^2}$ is finite on $[1,+\infty)$ while the integral of $\tfrac{1}{x}$ on the same interval isn't. If you plot both functions, you can get an intuitive feeling for this: if you look at the (unbounded) region between the function's graph and the $x$-axis, from $x=1$ but stretching to infinity, it can be possible to associate a finite number with this region which can be interpreted as the area. For this area to remain finite, the function needs to decrease (drop to $0$) "sufficiently fast". This is the case for $\tfrac{1}{x^2}$ but not for $\tfrac{1}{x}$.
Similarly, returning to your example, you can look at the graphs of $\tfrac{1}{x}$ and $\ln x$ (or $-\ln x$ for an easier comparison in the first quadrant). On $(0,1]$, there is an unbounded region between the graph, the $x$-axis and the $y$-axis. It is clear from the graphs that the region corresponding to $-\ln x$ is smaller than the region corresponding to $\tfrac{1}{x}$. When it is possible to associate a finite area with such a region, the corresponding improper integral converges. 
Although you cannot tell from looking at the graphs if the integrals will converge or not, it is clear that if only one converges, it'll be the one of the function $\ln x$ (or $-\ln x$ in the linked plot).
To check whether or not such an improper integral exists (i.e. converges), you use the definition and calculate the required limit:
$$\int_0^1 \ln x \,\mbox{d}x = \lim_{a \to 0^+} \int_a^1 \ln x \,\mbox{d}x = -1$$
while
$$\int_0^1 \frac{1}{x} \,\mbox{d}x = \lim_{a \to 0^+} \int_a^1 \frac{1}{x} \,\mbox{d}x = +\infty$$
A: It's not really that strange. The same behavior can be seen for the function $x\mapsto x^{-1/2}$, which is integrable over $(0,1]$, but its derivative isn't.
Whenever you have a continuous function $f$ on $[0,1]$ which is twice differentiable on $(0,1]$ and
$$
\lim_{x\to0}f'(x)=\pm\infty
$$
the same situation can appear: $f'(x)$ is certainly integrable over $(0,1]$, but $f''(x)$ may not be.
In the $\ln x$ case, the function $f$ is
$$
\begin{cases}
x(\ln x-1) & \text{if $0<x\le1$}\\[4px]
0 & \text{if $x=0$}
\end{cases}
$$
The fact that $\ln x$ has an antiderivative that can be extended by continuity at $0$ implies that
$$
\int_0^1 \ln x\,dx
$$
converges; on the other hand,
$$
\int_0^1 \frac{1}{x}\,dx
$$
does not converge.
Similarly, for $x\mapsto x^{-1/2}$, the function $f$ is $f(x)=2\sqrt{x}$.
You may try your hand into finding a function $f$ satisfying the above requirements and such that both $f'$ and $f''$ are integrable over $(0,1]$.
