What is $\int_0^1\frac{x^7-1}{\log(x)}\mathrm dx$?

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{\pp\pars{\mu} \equiv \int_{0}^{1}{x^{\mu} - 1 \over \ln\pars{x}}\,\dd x}$

$$ \pp'\pars{\mu} \equiv \int_{0}^{1}{x^{\mu}\ln\pars{x} \over \ln\pars{x}}\,\dd x = \int_{0}^{1}x^{\mu}\,\dd x = {1 \over \mu + 1} \quad\imp\quad \pp\pars{\mu} - \overbrace{\pp\pars{0}}^{=\ 0} = \ln\pars{\mu + 1} $$

$$ \pp\pars{7} = \color{#0000ff}{\large\int_{0}^{1}{x^{7} - 1 \over \ln\pars{x}} \,\dd x} = \ln\pars{7 + 1} = \ln\pars{8} = \color{#0000ff}{\large 3\ln\pars{2}} $$

Change of variables $\log(x) = -t$ makes this into $$ \int_0^\infty \dfrac{1 - e^{-7t}}{t} e^{-t}\ dt $$ More generally, for $\alpha \ge 0$ let $$f(\alpha) = \int_0^\infty \dfrac{1-\exp(-\alpha t)}{t} e^{-t}\ dt$$ Then $f(0) = 0$ while $$f'(\alpha) = \int_0^\infty \exp(-(\alpha+1) t)\ dt = \dfrac{1}{1+\alpha}$$ from which $$f(\alpha) = \ln(1+\alpha)$$

Yet another direct way forward is to use Frullani's Integral. To that end, let $I(a)$ be the integral given by

$$I(a)=\int_0^1 \frac{x^a-1}{\log x}\,dx$$

Enforcing the substitution $\log x \to -x$ yields

$$\begin{align} I(a)&=\int_{0}^{\infty} \frac{e^{-ax}-1}{x}\,e^{-x}\,dx\\\\ &=-\int_{0}^{\infty} \frac{e^{-(a+1)x}-e^{-x}}{x}\,dx \end{align}$$

whereupon using Frullani's Integral we obtain

$$\bbox[5px,border:2px solid #C0A000]{I=\log(a+1)}$$

For $a=7$, we have $$\bbox[5px,border:2px solid #C0A000]{I(7)=\log (8)}$$as expected!