Why does $\int^{ab}_{a} \frac{1}{x} dx = \int^{b}_{1} \frac{1}{t} dt$? I can't understand how the integral having limits from $a$ to $ab$ in Step 1 is equivalent to the integral having limits from $1$ to $b$.  I'm a beginner here. Please explain in detail.
\begin{align*}
\ln(ab) = \int^{ab}_{1} \frac{1}{x} dx &= \int^{a}_{1} \frac{1}{x} dx + \int^{ab}_{a} \frac{1}{x} dx\\
&= \int^{a}_{1} \frac{1}{x} dx + \int^{b}_{1} \frac{1}{at} d(at)\\
&= \int^{a}_{1} \frac{1}{x} dx + \int^{b}_{1} \frac{1}{t} dt\\
&= \ln(a) + \ln(b).
\end{align*}

Given that the integral $\int_a^b f(x)\>dx$ is "the area under the curve $y=f(x)$ for $x$ between $a$ and $b$", the equality of the two integrals
$$\int_1^b{dx\over x},\quad \int_a^{ab} {dt\over t}$$
follows with an elementary geometric argument: The map $$(x,y)\mapsto\left(a x,\>{y\over a}\right)$$
which stretches by the factor $a>0$ in $x$-direction and compresses by the same factor in $y$-direction maps the first area onto the second.
