How to calculate the integral $I=\int\limits_0^1\frac{x^n-1}{\ln(x)} \,\mathrm dx$ How can we calculate this integral:

$$I=\int\limits_0^1\frac{x^n-1}{\ln(x)}\,\mathrm dx$$

I believe that integral is equal to $\ln(n+1)$, but I don't lnow how to prove it.

We have
$$x^a = e^{a\log x} = \sum_{k=0}^{\infty} \dfrac{(a \log x)^k}{k!}$$
Hence,
$$\dfrac{x^a-1}{\log(x)} = \sum_{k=1}^{\infty} \dfrac{a^k \log^{k-1}(x)}{k!}$$
Make use of the  fact that
$$\int_0^1 \log^m(x) dx = (-1)^m m!$$
Hence, we obtain
$$\dfrac{x^a-1}{\log(x)} = \sum_{k=1}^{\infty} \dfrac{a^k}{k!} (-1)^{k-1} (k-1)! = \sum_{k=1}^{\infty} (-1)^{k-1} \dfrac{a^k}k =  \log(1+a)$$
Note that the above derivation is only valid when $a \in (-1,1)$. For $a\geq1$, it is easy to show by similar method that
$$I(a) - I(a-1) = \log(1+1/a)$$
Since $I(a) = \log(1+a)$ on the interval $[0,1]$, we can now conclude that
$$I(a) = \log(1+a)$$
for all $a \geq -1$.
