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}
 A: 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.
A: I think this way will be better. As your solution, my solution also uses power series expansion:
By integration by parts, we see that
$$a_n=\ln2-\int_{0}^1\frac{nx^n}{1+x^n}=\ln2-n(1-\int_0^1\frac{dx}{1+x^n}).$$
Now, note that
$$n(1-\int_0^1\frac{dx}{1+x^n})=n(1-\int_0^1\sum_{k=0}^{\infty}(-1)^kx^{nk})\\
=n(1-\sum_{k=0}^{\infty}\frac{(-1)^k}{nk+1})\\
=\sum_{k=1}^{\infty}\frac{(-1)^kn}{nk+1}$$
Now:
$$\lim_{n\to\infty}a_n=\ln2-\lim_{n\to\infty}\sum_{k=1}^{\infty}\frac{(-1)^kn}{nk+1}\\
=\ln2-\sum_{k=1}^\infty\frac{(-1)^k}{k}=0$$
A: Integrate by parts
$$a_n=\int_0^1 \ln({x^n+1})dx= -n+\ln2+ \int_0^1 \frac n{x^n+1}dx$$
With the roots
$x_k= e^{i\frac{(2k-1)\pi}n} $ for $x^n+1=0$,  integrate
\begin{align}
\int_0^1 \frac{n}{x^n+1}dx & = -\sum_{k=1}^n\int_0^1 \frac{x_k}{x-x_k} dx
=  -\sum_{k=1}^nx_k\ln(1-x_k^{-1})\\
 &= -\sum_{k=1}^n x_k \left[i \frac{n-2k+1}{2n}+ \ln \left(2 \sin\frac{(2k-1)\pi}{2n}\right) \right]
\end{align}
Per $\sum_{k=1}^nx^k=0$ and the symmetry of $x^k$ to arrive at the close-form
$$\int_0^1 \frac n{x^n+1}dx =\sum_{k=1}^{[\frac n2]} ( \pi-\theta_k )\sin\theta_k -\cos\theta_k \ln\sin^2\frac{\theta_k}2,\>\>\>
\theta_k= \frac{2k-1}{n} \pi$$
Listed below are the results for the first few $n’s$
\begin{align}
& a_2 = - 2 + \ln 2 + \frac\pi2\\
 & a_3 = -3 +2\ln2 + \frac\pi{\sqrt3}\\
 & a_4 = -4 +\ln 2+ \frac\pi{\sqrt2}+\sqrt2\ln(1+\sqrt2)\\
 & a_5 = -5 +2\ln2 +\frac\pi{\sqrt{10}}\sqrt{5+\sqrt5}+\sqrt5\ln\frac{1+\sqrt5}2\\
 & a_6 =-6+\ln2 + \pi +\sqrt3\ln(2+\sqrt3) \\
 & a_7 =-7+\ln2 + \frac\pi2\csc\frac\pi7 
+\frac{\ln\csc^2\frac\pi{14}}{\sec\frac\pi7} 
+\frac{\ln\csc^2\frac{3\pi}{14}}{\sec\frac{3\pi}7}
+\frac{\ln\csc^2\frac{5\pi}{14}}{\sec\frac{5\pi}7}
\end{align}
A: From where you stopped $$\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \frac{1}{(n+1)(k+1)-k},$$ if you multiply out the denominator, you see that it can be easily factored, so that we have $$\frac 1n\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)(k+1+\frac 1n)},$$ which after performing the shift $k\mapsto k-1$ becomes $$\frac 1n\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k(k+\frac 1n)},$$ which should have a closed form, since the series is very similar to $$\sum_{k=1}^\infty \frac{\frac 1n}{k(k+\frac 1n)},$$ which has a closed form.

See, for example, what Wolfram Alpha gives, but I think there is an elementary representation since the last series above also has an elementary closed format. I shall attempt to look for it; if I find it (before you or someone else, of course) I'll edit this to let you know.
