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

/A problem from the 2012 MIT Integration Bee is
$$
\int_0^1\frac{x^7-1}{\log(x)}\mathrm dx
$$
The answer is $\log(8)$. Wolfram Alpha gives an indefinite form in terms of the logarithmic integral function, but times out doing the computation. Is there a way to do it by hand?

Answer

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

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

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

Attribution
Source : Link , Question Author : YoniY , Answer Author : Felix Marin

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