Series expansion of $ \int_{0}^{1} \frac{x^n}{1+x^n} \mathrm dx $ I have proved $$ \int_{0}^{1} \frac{x^n}{1+x^n} \mathrm dx \sim \frac{\ln(2)}{n}$$ 
How can I get further and find $ a$ such that:
$$ \int_{0}^{1} \frac{x^n}{1+x^n} \mathrm dx=\frac{\ln(2)}{n}+\frac{a}{n^2}+o\left(\frac{1}{n^2}\right)? $$
 A: $$
\begin{eqnarray}
\int_{0}^{1} \frac{x^n}{1+x^n} \mathrm dx
&=&
-\int_{0}^{1} \sum_{k=1}^\infty(-x^n)^k \mathrm dx
\\
&=&
-\sum_{k=1}^\infty\int_{0}^{1}(-x^n)^k \mathrm dx
\\
&=&
-\sum_{k=1}^\infty\frac{(-1)^k}{kn+1}
\\
&=&
-\sum_{k=1}^\infty\frac1{kn}\frac{(-1)^k}{1+\frac1{kn}}
\\
&=&
\sum_{k=1}^\infty\sum_{l=0}^\infty(-1)^k(-kn)^{-(l+1)}
\\
&=&
-\sum_{l=0}^\infty (-1)^ln^{-(l+1)}\sum_{k=1}^\infty(-1)^kk^{-(l+1)}
\\
&=&
-\frac1n\sum_{k=1}^\infty\frac{(-1)^k}k+\sum_{l=1}^\infty(-1)^ln^{-(l+1)}\left(1-2^{-l}\right)\zeta(l+1)
\\
&=&
\frac{\log2}n-\frac{\pi^2}{12}\frac1{n^2}+o\left(\frac1{n^2}\right)\;.
\end{eqnarray}
$$
A: Put $I-n:=\int_0^1\frac{x^n}{1+x^n}$, and make the substitution $t=x^n$ to get 
\begin{align*}
I_n&=\frac 1n\int_0^1\frac{t^{\frac 1n}}{1+t}dt\\
&=\frac 1n\left[\ln(1+t)t^{\frac 1n}\right]_0^1-\frac 1n\int_0^1\ln(1+t)\frac 1nt^{\frac 1n-1}dt\\
&=\frac{\ln 2}n-\frac 1{n^2}\int_0^1\frac{\ln(1+t)}tdt+\frac 1{n^2}\int_0^1\frac{\ln(1+t)}t(1-t^{\frac 1n})dt. 
\end{align*} 
Since $|1-t^{\frac 1n}|=\left|\int_0^{\frac{\ln t}n}e^t dt\right|\leq \frac 1n|\ln t|$, we have $\lim_{n\to\infty}\int_0^1\frac{\ln(1+t)}t(1-t^{\frac 1n})dt=0$
hence $a=-\int_0^1\frac{\ln(1+t)}tdt$. We can compute this integral using power series.
