Proving an integration equality I am interested in why $$\int_0^1\dfrac{\ln(1+x^n)}{x}dx=\frac{\pi^2}{12(n)}$$
This is what WA gives me http://www.wolframalpha.com/input/?i=integral+of+ln%281%2Bx%5En%29%2Fx
Is there a way to prove this?
 A: Enforce the substitution $x\to x^{1/n}$ to find
$$\begin{align}
\int_0^1 \frac{\log(1+x^n)}{x}\,dx&=\frac1n \int_0^1 \frac{\log(1+x)}{x}\,dx \tag 1\\\\
&=\frac1n \left.\left(-\text{Li}_2(-x)\right)\right|_0^1\\\\
&=-\frac1n \text{Li}_2(-1)\\\\
&=\frac{\pi^2}{12n}
\end{align}$$

NOTE:
If one is unfamiliar with the dilogarithm function $\text{Li}_2(z)$, we can simply write the right-hand side of $(1)$ as
$$\begin{align}
\frac1n \int_0^1 \frac{\log(1+x)}{x}\,dx &=\frac1n \int_0^1 \frac{\sum_{k=1}^\infty \frac{(-1)^{k-1} x^k}{k}}{x}\,dx\\\\
&=\frac1n \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^2}\\\\
&=\frac{\pi^2}{12n}
\end{align}$$
A: First, note the following:
$$\ln(1+x)=\sum_{k=1}^{\infty}\frac{x^k(-1)^{k+1}}{k}\implies\ln(1+x^n)=\sum_{k=1}^{\infty}\frac{x^{kn}(-1)^{k+1}}{k}$$
Now, since our limits are from $0$ to $1$ we are fine to proceed with integrating. Therefore, we now have:
$$\int_0^1 \sum_{k=1}^{\infty}\frac{x^{kn}(-1)^{k+1}}{k}\,dx=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\int_0^1x^{kn-1}\,dx=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{nk^2}$$
Now, note that:
$$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}=\sum_{k=1}^\infty \frac{1}{k^2}-\sum_{k=1}^\infty \frac{2}{(2k)^2}=\sum_{k=1}^\infty \frac{1}{k^2}-\sum_{k=1}^\infty \frac{1}{2k^2}=\frac12\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{12}$$
$$\therefore \int_0^1\frac{\ln(1+x^n)}{x}\,dx=\frac{\pi^2}{12n}$$
