$\log$ as $\int 1/x$
this actually isn't so insane from looking at the limits of the graphs:
$\int_1^x t^{s-1} = \frac{1}{s} (x^s - 1)$
for $s>0$
you get stuff like $n (x^{1/n} - 1)$,
so a root that keeps elongating at $x\to 0$
and flattening at $x\to\infty$

[Note that for all of these, you get
 $f(1) = 0$  (b/c it's $\int_1^x ...$)
 $f'(1) = 1$ (b/c it's $x^p$]

for $s<0$
you get stuff like
$n (1 - 1/x^n)$
so a flipped asymptote that keeps elongating at $x\to \infty$
(so the asymptote diverges to infinity)
...and slowing down at $x\to 0$

...so these do really come together.

This looks -much- better if you graph it
(which I haven't, but can visualize, natch)

Algebraically, this is also understandable:
We're saying:
$\log x = \lim_{n\to \infty} n (x^{1/n} - 1)$
Contrast:
$\exp x = \lim_{n\to \infty} (1+x/n)^n$
Unsurprisingly, these are inverses.

Stated -this- way, it's algebraically obvious
(grrr...and no-one ever told me this.)


That all said, it's still quite weird that powers
are closed under derivatives but not under integrals,
and that this weird one should show up.
One way to look at this is:
let $E$ be the span of $x^s$ as $s \in \bR$
Then $D\from E \to E$ has a kernel, namely (span of) $1=x^0$,
and a cokernel;
  unsurprisingly, as span $x^s$ goes to span $x^{s-1}$,
  so you exactly miss $x^{0-1}=x^{-1}$
Integration always gives you some function
  (esp. since these are continuous (away from 0))

oh, and if you iterate this you just get more logs:
$\int_1^x \log = x \log x - x$
  (see cute disection proof elsewhere)
