Finding the integral $\int_0^1 \frac{x^a - 1}{\log x} dx$ How to do the following integral:
$$\int_{0}^1 \dfrac{x^a-1}{\log(x)}dx$$
where $a \geq 0$?
I was asked this question by a friend, and couldn't think of any substitution that works. Plugging in a=2,3, etc in Wolfram, I get values like $\log(a+1)$, which may be the right answer (for general $a$). Is there a simple way to calculate this integral?

We can utilize
$$
\int_0^1x^t\,\mathrm{d}t=\frac{x-1}{\log(x)}
$$
combined with the substitution $x\mapsto x^{1/a}$, to get
$$
\begin{align}
\int_0^1\frac{x^a-1}{\log(x)}\,\mathrm{d}x
&=\int_0^1\frac{x-1}{\log(x)}x^{\frac1a-1}\,\mathrm{d}x\\
&=\int_0^1\int_0^1x^{\frac1a-1}x^t\,\mathrm{d}t\,\mathrm{d}x\\
&=\int_0^1\int_0^1x^{\frac1a-1}x^t\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_0^1\frac1{\frac1a+t}\,\mathrm{d}t\\
&=\log\left(\frac1a+1\right)-\log\left(\frac1a\right)\\[9pt]
&=\log(1+a)
\end{align}
$$
