Let$$L(x)=\int_{1}^{x}\frac{1}{t}dt.$$How can I show that $L(ab)=L(a)+L(b)?$ This is what I have tried: If we let$$\begin{align}L(ax)&=\int_{1}^{ax}\frac{1}{t}dt,\text{ and}\\L(x)&=\int_{1}^{x}\frac{1}{t}dt,\end{align}$$then, by the FTC,$$\begin{align}F(ax)-F(1)&\Longrightarrow aF^{'}(ax)=a\frac{1}{ax}=\frac{1}{x},\text{ and}\\F(x)-F(1)&\Longrightarrow F^{'}(x)=\frac{1}{x}.\end{align}$$Hence, $L(ax)=L(x)+C$, where $C$ is some constant, and it must be $L(a)$.
Is this enough to show that $L(ab)=L(a)+L(b)$? How could I go about doing the same for $L(a^r)=rL(a)$?