Why, intuitively, does $log(x)$ come in as the integral of $1/x$, wheras the integral of other powers of $x$ are powers of $x$? Question in title really, something I always found strange when I was learning calculus.
I can see that $\int \frac{1}{x} dx$ can't be $\frac{x^0}{0}$ since this is not defined, and then the definite integral $\int_1^t \frac{1}{x} dx$ comes down to $$\lim_{\delta \rightarrow 0} \frac{t^{\delta}-1}{\delta} = \log (t).$$ 
But this understanding just comes from l'Hopital's rule, and also it still just seems really bizaare that the log function should fit into the set of power functions like this.  Can anyone de-mystify this at all?
 A: It is beautiful that this works out so nicely.  
Here's a thought.  The key property of $\log$ is that it's the inverse of $e^x$, and the key property of $e^x$ is that $\frac{d}{dx} e^x = e^x$.  Because the derivative of $e^x$ is so simple, we might expect some correspondingly nice result for the derivative of its inverse function.
Differentiating both sides of
\begin{equation}
\log(e^x) = x \qquad (\spadesuit)
\end{equation}
yields
\begin{align}
& \log'(e^x) e^x = 1 \\
\implies & \log'(y) = \frac{1}{y} \quad \text{for all } y > 0.
\end{align}
So, the fact that the derivative of $\log(y)$ is $\frac{1}{y}$
is almost just a restatement of $(\spadesuit)$.
A: Note the limit you wrote doesn't exist: it's of the form $1/0$.
However, let's instead consider the definite integral
$$ \int_1^x t^{a-1} \, dt = \frac{x^a - 1}{a}  \qquad \qquad a \neq 0 $$
Taking the limit of this as $a \to 0$ we can do, and we can rationalize this being the logarithm as follows:
$$\begin{align} f(xy) = \lim_{a \to 0} \frac{(xy)^a - 1}{a} &= \lim_{a \to 0} \frac{x^a (y^a - 1) + (x^a - 1)}{a} 
\\&= \lim_{a \to 0} x^a \frac{y^a - 1}{a} + \lim_{a \to 0} \frac{x^a - 1}{a} 
\\&= 1 \cdot f(y) + f(x)
\\&= f(x) + f(y)
\end{align}$$
so it ought to satisfy the right algebraic identity. In fact, there's even a theorem that says any sufficiently nice function (I think continuous is enough) that satisfies $f(xy) = f(x) + f(y)$ must be a logarithm.
