Proving properties of logarithm using integral form

Question

Let: $$L(x) = \int_1^x\frac{1}{u}du$$ we are trying to show that $L(x^b) = bL(x)$

I was trying: $$L(x^b) = \int_1^{x^{b}}\frac{1}{u}du = \int_1^{x}\frac{1}{u}du + \int_x^{x^{b}}\frac{1}{u}du$$

Let's focus on the last term using $u = bt$ and $dt = bdt$:

$$\int_x^{x^{b}}\frac{1}{u}du = \int_{ t =\frac{x}{b}}^{ t =\frac{x^b}{b}}\frac{1}{bt}bdt$$

but this did not work out in showing the last form has the correct form

Answer 1

To prove $$\int_1^{x^b}{\frac{1}{u} du}=b\int_1^{x}{\frac{1}{u} du} $$

we put $u=v^b$,

then $du= b v^{b-1} dv$,

$u=1 \implies v=1$ and $u=x^b \implies v=x $

$\therefore$ \begin{align} \int_1^{x^b}{\frac{1}{u} du} & = \int_1^{x}{\frac{bv^{b-1}}{v^b} dv} \\ & = b\int_1^{x}{\frac{1}{v} dv} \\ & = b\int_1^{x}{\frac{1}{u} du} \\ \end{align}

Answer 2

Using $s=u/x$ then $xds=du$

$$\int_x^{x^{b}}\frac{1}{u}du = \int_1^{x^{b-1}}\frac{1}{sx}xds=\int_1^{x^{b-1}}\frac{1}{s}ds$$