1. ## Prove directly

How to prove the following equality directly without assuming it represents the natural log function?
$\displaystyle \int_{1}^{yz} t^{-1}\, dt=\int_{1}^{y} t^{-1}\, dt+\int_{1}^{z} t^{-1}\, dt.$
Thanks!

2. Proposition:

$\displaystyle{\int_1^x \frac{1}{t}dt = \int_a^{ax}\frac{1}{t}dt}$

Proof: Let $u=at\Rightarrow dt=\frac{du}{a}$. If $t=1$, then $u=a$ and if $t=x$, $u=ax$. Therefore:

$\displaystyle{\int_1^x\frac{1}{t}dt=\int_a^{ax}\fr ac{1}{u/a}\cdot\frac{du}{a} = \int_a^{ax}\frac{1}{u}du }$

This show us that this integral is invariant by translation.

So $\displaystyle{\int_1^{yz} \frac{1}{t}dt = \int_1^{y}\frac{1}{t}dt + \int_y^{yz} \frac{1}{t}dt}$.

By the proposition you get the answer.