Closed Form of $a_n = \int_0^1 \ln(1+x^n) dx$ I want to know the closed form of :
$$a_n = \int_0^1 \ln(1+x^n)dx, \quad \forall n \in \mathbb{N}$$

I found : 
$$0<a_n<\frac{1}{n+1}, \quad \lim_{n\to\infty} a_n =0$$
I started from
\begin{align}
&\ln (1+x)=\sum_{k=0}^\infty \frac{(-1)^k x^{k+1}}{k+1} \\
& \Rightarrow \ln(1+x^n)=\sum_{k=0}^\infty \frac{(-1)^k x^{nk+n}}{k+1} \\
& a_n = \int_0^1 \ln(1+x^n)dx = \sum_{k=0}^\infty \frac{(-1)^k}{k+1}\int_0^1 x^{n+nk}dx \\
&=\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \times \frac{1}{n+nk+1} \\
&=\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \times \frac{1}{(n+1)(k+1)-k} 
\end{align}
But I stucked here. Is there any closed (or approximated) from exist?

These are some results for litte $n$ : 
\begin{align}
& a_1 = 2\ln 2 - 1 \\
& a_2 = \ln2 - 2 + \frac{\pi}{2} \\
& a_3 = 2\ln 2 - 3 + \frac{\pi \sqrt{3}}{3} \\
\end{align}

We have
\begin{align} \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)(k+z)} &= \sum_{k=0}^\infty \frac{1}{(k+1)(k+z)} - \sum_{k=0}^\infty \frac{1-(-1)^k}{(k+1)(k+z)} = \\
&= \sum_{k=0}^\infty \frac{1}{(k+1)(k+z)} - \sum_{m=0}^\infty \frac{2}{(2m+2)(2m+1+z)} = \\
&= \sum_{k=0}^\infty \frac{1}{(k+1)(k+z)} - \frac12 \sum_{m=0}^\infty \frac{1}{(m+1)(m+\frac{1+z}{2})} = \\
&= \frac{\psi(z)+\gamma}{z-1} - \frac12 \frac{\psi(\frac{1+z}{2})+\gamma}{\frac{1+z}{2}-1} = \\
&= \frac{\psi(z)-\psi(\frac{1+z}{2})}{z-1}\end{align}
where $\psi(z)$ is the digamma function and $\gamma$ is Euler-Mascheroni constant.
We have then
\begin{align} \int_0^1 \ln(1+x^n)dx &=\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)(n+nk+1)} = \\
&= \frac{1}{n} \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)(k+1+\frac{1}{n})} = \\&= \frac{1}{n} \frac{\psi(1+\frac{1}{n})-\psi(1+\frac{1}{2n})}{(1+\frac{1}{n})-1} = \\
&= \psi\big(1+\frac{1}{n}\big)-\psi\big(1+\frac{1}{2n}\big)\end{align}
It turns out that for any $n\in\mathbb N$ it can be expressed in terms of elementary functions. This is because
\begin{align} \psi\big(1+\frac{1}{n}\big) &= -\gamma + \sum_{k=0}^\infty\frac{\frac1n}{(k+1)(k+1+\frac1n)} = \\
&= -\gamma + \sum_{k=0}^\infty\Big(\frac{1}{k+1}-\frac{1}{k+1+\frac1n}\Big) = \\
&= -\gamma + \sum_{k=0}^\infty\int_0^1 (x^k-x^{k+\frac1n})dx = \\
&= -\gamma + \int_0^1 \frac{1-x^{\frac1n}}{1-x}dx = \\
&= -\gamma + \int_0^1 \frac{(1-y)ny^{n-1}}{1-y^n}dy = \\
&= -\gamma + \int_0^1 \frac{ny^{n-1}}{\prod_{k=1}^{n-1}(y-e^{i\frac{2\pi k }{n}})}dy \end{align}
and the last integral can be calculated using the partial fraction decomposition.
