<s>[INST] 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}
 [/INST] 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:
$$\lim_{n\to\infty}a_n=\ln2-\lim_{n\to\infty}\sum_{k=1}^{\infty}\frac{(-1)^kn}{nk+1}\\
=\ln2-\sum_{k=1}^\infty\frac{(-1)^k}{k}=0$$
</s>