What is so special about α=−1\alpha=-1 in the integral of xαx^\alpha?

Of course, it is easy to see, that the integral (or the
antiderivative) of $f(x) = 1/x$ is $\log(|x|)$ and of course for
$\alpha\neq - 1$ the antiderivative of $f(x) = x^\alpha$ is
$x^{\alpha+1}/(\alpha+1)$.

I was wondering if there is an intuitive (probably geometric)
explanation why the case $\alpha=-1$ is so different and why the
logarithm appears?

Some answers which I thought of but which are not convincing:

1. Taking the limit $\alpha=-1$ either from above or below lead to diverging functions.

2. Some speciality of the case $\alpha=-1$ are that both asymptotes are non-integrable. However, the antidrivative is a local thing, and hence, shouldn’t care about the behavior at infinity.

Assume you know that for every $\beta$ the derivative of the function $x\mapsto x^\beta$ is the function $x\mapsto\beta x^{\beta-1}$ and that you want to choose $\beta$ such that the derivative is a multiple of the function $x\mapsto x^{\alpha}$. You are led to solve the equation $\beta-1=\alpha$, which yields $\beta=\alpha+1$. If $\alpha=-1$, this gets you $\beta=0$, but then the derivative you obtain is the function $x\mapsto 0x^{-1}=0$, which is not a nonzero multiple of $x\mapsto x^{-1}$. For every other $\alpha$, this procedure gets you an antiderivative but for $\alpha=-1$, you failed. Or rather, you proved that no power function is an antiderivative of $x\mapsto x^{-1}$. Your next step might be (as mathematicians often do when they want to transform one of their failures into a success) to introduce a new function defined as the antiderivative of $x\mapsto x^{-1}$ which is zero at $x=1$, and maybe to give it a cute name like logarithm, and then, who knows, to start studying its properties…
Fix $s>t>0$ and $c>1$ and consider the area under the curve $x\mapsto x^\alpha$ between the abscissæ $x=t$ and $x=s$ on the one hand and between the abscissæ $x=ct$ and $x=cs$ on the other hand. Replacing $x$ by $cx$ multiplies the function by a factor $c^\alpha$. The length of the interval of integration is multiplied by $c$ hence the integral itself is multiplied by $c^{\alpha+1}$.
On the other hand, if an antiderivative $F$ of $x\mapsto x^\alpha$ is a multiple of $x\mapsto x^\beta$ for a given $\beta$, then $F(ct)=c^\beta F(t)$ and $F(cs)=c^\beta F(s)$ hence $F(ct)-F(cs)=c^\beta (F(t)-F(s))$. Note that this last relation holds even if one assumes only that $F$ is the sum of a constant and a multiple of $x\mapsto x^\beta$.
Putting the two parts together yields $c^{\alpha+1}=c^\beta$. Once again, if $\alpha=-1$, this would yield $\beta=0$, hence $F$ would be constant and the area $F(t)-F(s)$ under the curve $x\mapsto x^\alpha$ from $s$ to $t\ge s$ would be zero for every such $s$ and $t$, which is impossible since the function $x\mapsto x^\alpha$ is not zero. (And for every $\alpha\ne1$, this scaling argument yields the correct exponent $\beta$.)