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*}
 A: The substitution method was already given so take this method:
Let 
$$f(x)=\int_1^x\frac{dt}t$$
and 
$$g(x)=\int_a^{ax}\frac{dt}{t}$$
then we have
$$f'(x)=g'(x)=\frac1x$$
and
$$f(1)=g(1)=0$$
hence we conclude that
$$f(x)=g(x),\quad \forall\; x\in\Bbb R$$
A: 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.
A: 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.
