I'm stuck trying to evaluate the following integral:
$$ \int_0^1 \frac{x^7-1}{\log(x)}dx $$
This is what I have tried. Substituting $u = \log(x)$ gives $x = e^u$ and $dx = x\cdot du = e^u du$, and thus
$$ \int_{-\infty}^0 \frac{e^{7u} - 1}{u}e^udu $$
which simplifies to
$$ \int_{-\infty}^0 \frac{e^{8u}}{u} - \frac{e^u}{u}du. $$
It would suffice to know the antiderivative of $e^x / x$, but unfortunately, it doesn't have a closed form. How can I proceed from here?
