<s>[INST] 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*}
 [/INST] Your are asking why 
$$ \int_{a}^{ab} \frac{1}{x} dx = \int_{1}^b \frac{1}{t} dt $$
Well, it follows by substituting $x = at \implies dx = a dt $.
Now, the limits in the first integral are $x =a$ to $x=ab$. Hence, if $x =a $, then $ t = \frac{a}{a} = 1 $. and if $ x = ab $ , then $ t = \frac{ab}{a} = b$ which are your new limits of integration.
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