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.

Let: $$I(n)=\int_0^1\dfrac{x^n-1}{\ln x}dx$$
Then: \begin{align}I'(n)&=\dfrac{d}{dn}\int_0^1\dfrac{x^n-1}{\ln x}dx=\int_0^1\dfrac{\partial}{\partial n}\left[\dfrac{x^n-1}{\ln x}\right]dx\\
&=\int_0^1 x^n dx=\left.\dfrac{x^n+1}{n+1}\right\vert_0^1\\
&=\dfrac{1}{n+1} \end{align}
Therefore: $$I(n)=\int I'(n)=\ln(n+1)$$
