Let $x$ be a real number and $z \neq 0$ a complex number, the following can be proved using the Taylor series of a power.
$$\int x^z = \frac{1}zx^{z-1}$$
Do you know any more elementary solution?
Let $x$ be a real number and $z \neq 0$ a complex number, the following can be proved using the Taylor series of a power.
$$\int x^z = \frac{1}zx^{z-1}$$
Do you know any more elementary solution?
Let $x^z=\exp(z\ln(x))$. It is then not too hard to see that
$$\frac d{dx}\exp(z\ln(x))=\frac zx\exp(z\ln(x))=zx^{z-1}$$
which is the power rule. In reverse, we get the antiderivative:
$$\int x^z\ dx=\frac{x^{z+1}}{z+1}+c$$
For the case when $z=-1$, the antiderivative may be handled using this answer of mine, which follows through power rule.
This is no more complicated than just defining the complex derivative as
$$\lim_{h \to 0} \frac{f(z+h)-f(z)}{h}$$
Where both $h$ and $z$ are in $\mathbb{C}$. Now we just note that
$$\frac{d}{dz} \frac{x^{z+1}}{z+1} = x^z$$
and we define the complex antiderivative as the operator that inverts complex differentiation.