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?

I thought it might be instructive to present yet another approach. 
Note that using $\int_0^1 x^t \,dt=\frac{x-1}{\log(x)}$ we can write
$$\begin{align}
\int_0^1 \frac{x^7-1}{\log(x)}\,dx&=\int_0^1 (x^6+x^5+x^4+x^3+x^2+x+1)\left(\int_0^1 x^t\,dt\right)\,dx\tag1\\\\
&=\int_0^1\int_0^1 (x^{t+6}+x^{t+5}+x^{t+4}+x^{t+3}+x^{t+2}+x^{t+1}+x^t)\,dx\,dt\tag2\\\\
&=\int_0^1 \left(\frac{1}{t+7}+\frac{1}{t+6}+\frac{1}{t+5}+\frac{1}{t+4}+\frac{1}{t+3}+\frac{1}{t+2}+\frac{1}{t+1}\right)\,dt\\\\
&=\log(8)
\end{align}$$
as expected, where the Fubini-Tonelli Theorem guarantees the legitimacy of interchanging the order of integration in going from $(1)$ to $(2)$.

Note that if we first enforce the substitution $x^7\to x$, we obtain
$$\begin{align}
\int_0^1 \frac{x^7-1}{\log(x)}\,dx&=\int_0^1 x^{1/7-1}\left(\frac{(x-1)}{\log(x)}\right)\,dx\\\\
&=\int_0^1 x^{1/7-1}\left(\int_0^1 x^t\,dt\right)\,dx\\\\
&=\int_0^1\int_0^1 x^{t+1/7-1} \,dx\,dt\\\\
&=\int_0^1 \frac{1}{t+1/7}\,dt\\\\
&=\log(8)
\end{align}$$
as expected!
