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.
 A: One may observe that $$\frac{x^n-1}{\ln x}=n\int_{0}^{1}x^{nu} du$$ plugging it in your initial integral, applying Fubini's theorem, gives
$$
\int_0^1\frac{x^n-1}{\ln x}dx= n\int_0^1\!\!\frac{1}{nu+1}du=\int_0^1\frac{(nu+1)'}{nu+1}du=\ln(n+1).
$$
A: If we replace $x$ with $e^{-t}$ we are left with:
$$ I = \int_{0}^{+\infty}\frac{e^{-t}-e^{-(n+1)t}}{t}\,dt = \log(n+1) $$
by Frullani's theorem.
A: 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: $$I(n)=\int I'(n)=\ln(n+1)$$
A: Hint: $(a^x)'=a^x\cdot log(a)$
A: 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 $a \geq -1$.
