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 have $x^a-1 = e^{a\log(x)}-1$. Hence, the integral is
\begin{align}
I & = \int_0^1 \dfrac{x^a-1}{\log(x)}dx = \int_0^1 \left(\sum_{k=1}^{\infty} \dfrac{a^k \log^k(x)}{k!}\right)\dfrac{dx}{\log(x)} = \sum_{k=1}^{\infty} \dfrac{a^k}{k!} \int_0^1 \log^{k-1}(x)dx\\
& = \sum_{k=1}^{\infty} \dfrac{a^k}{k!} (-1)^{k-1} (k-1)! = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}a^k}k = \log(1+a)
\end{align}
