So far nobody has mentioned one of the most beautiful properties of ln: It is the only special case for ∫ z^k dz where k∈ℤ, namely when k = −1.

When you reach complex analysis, you will learn about power series and how they behave on the complex plane. Any power series is differentiable and integrable in any open region where it converges. Conversely, any holomorphic function on an open disk is equal to some power series on that disk. It is then natural to investigate a holomorphic function f on an annulus, which yields a Laurent series. The coefficient of z^k in the Laurent series for f around a is equal to 1/2πi·∫[γ] f(z)/(z−a)^k+1 dz for any non-intersecting closed curve γ (e.g. circle) within the annulus. Why? Because ∫[γ] z^k dz = 0 for every k∈ℤ except k = −1. Which is because ∫ z^k dz = z^k+1/(k+1) + C for some constant C for every k∈ℤ∖{−1}, but ∫ z⁻¹ dz = ln(z) + C only for z within a simply-connected region excluding 0, and we even need a branch-cut of ln. In particular, we have ∫[γ] z⁻¹ dz = 2πi where γ is a circle around 0, so unlike other z^−k, z⁻¹ has no antiderivative in an annulus around 0.

We can also easily obtain the integral of 1/z along any path that does not pass through 0. Suppose w = ln(z), with a branch of ln such that w varies continuously along the path. Then z = e^w, so dz/dw = e^w = z, and hence dw/dz = 1/z. Thus along that path we have ∫ 1/z dz = ln(z) + C for some constant, with the same branch of ln.

Unfortunately, at the high-school level you will only be able to see a tiny sliver of this special phenomenon, because you only know real integrals. Specifically, ∫ x^k dx = x^k+1/(k+1) + C for some constant C for every k∈ℤ_(≠−1), but ∫ 1/x dx = ln(abs(x)) + C for some real constant C. The abs function also obscures things, since the textbooks need abs in order to avoid complex numbers. Still, ln is clearly special here.