Seeking for other methods to evaluate $\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx$ for $n\geq 2$. Inspired by my post, I go further to investigate the general integral and find a formula for
$$
I_n=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx =-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right] \tag*{}
$$
Let’s start with its partner integral
$$
I(a)=\int_0^{\infty}\left(x^n+1\right)^a d x
$$
and transform $I(a)$, by putting $y=\frac{1}{x^n+1}$, into a beta function
$$
\begin{aligned}
I(a) &=\frac{1}{n} \int_0^1 y^{-a-\frac{1}{n}-1}(1-y)^{-\frac{1}{n}-1} d y \\
&=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right)
\end{aligned}
$$
Differentiating $I(a)$ w.r.t. $a$ yields
$$
I^{\prime}(a)=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right)\left(\psi(-a)-\psi\left(-a-\frac{1}{n}\right)\right)
$$
Then putting $a=-1$ gives our integral$$
\begin{aligned}
I_n&=I^{\prime}(-1) \\&=\frac{1}{n} B\left(1-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(1)-\psi\left(1-\frac{1}{n}\right)\right] \\
&=-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right]\end{aligned}
$$

For examples,
$$
\begin{aligned}& I_2=-\frac{\pi}{2} \csc \frac{\pi}{2}\left[\gamma+\psi\left(1-\frac{1}{2}\right)\right]=\pi \ln 2,\\ & I_3=-\frac{\pi}{3} \csc \left(\frac{\pi}{3}\right)\left[\gamma+\psi\left(\frac{2}{3}\right)\right]=\frac{\pi \ln 3}{\sqrt{3}}-\frac{\pi^2}{9} ,\\ &I_4=-\frac{\pi}{4} \csc \left(\frac{\pi}{4}\right)\left[\gamma+\psi\left(\frac{3}{4}\right)\right]=\frac{3 \pi}{2 \sqrt{2}}\ln 2-\frac{\pi^2}{4 \sqrt{2}},\\
& I_5=-\frac{\pi}{5} \csc \left(\frac{\pi}{5}\right)\left[\gamma+\psi\left(\frac{4}{5}\right)\right]=-\frac{2 \sqrt{2} \pi}{5 \sqrt{5-\sqrt{5}}}\left[\gamma+\psi\left(\frac{4}{5}\right)\right], \\
& I_6=-\frac{\pi}{6} \csc \left(\frac{\pi}{6}\right)\left[\gamma+\psi\left(\frac{5}{6}\right)\right]=\frac{2 \pi}{3} \ln 2+\frac{\pi}{2} \ln 3-\frac{\pi^2}{2 \sqrt{3}},
\end{aligned}
$$

Furthermore, putting $a=-m$, gives
$$\boxed{I(m,n)=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{(x^n+1)^m} dx = \frac{1}{n} B\left(m-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(m)-\psi\left(m-\frac{1}{n}\right)\right] }$$
For example,
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(x^6+1\right)}{(x^6+1)^5} dx  & =\frac{1}{6} B\left(\frac{29}{6}, \frac{1}{6}\right)\left[\psi(5)-\psi\left(\frac{29}{6}\right)\right] \\
& =\frac{1}{6} \cdot \frac{21505 \pi}{15552} \cdot\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right) \\
& =\frac{21505 \pi}{93312}\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right)
\end{aligned}
$$

Are there any other methods?
Your comments and alternative methods are highly appreciated.
 A: The integral admits elementary close-form as evaluated below
\begin{align}
I_n=&\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} \overset{x\to 1/x}{dx}\\
=& \ \frac12 \int_0^{\infty} \frac{(1+x^{n-2})\ln \left(x^n+1\right)}{x^n+1}dx-\frac n2 \int_0^{\infty} \frac{x^{n-2}\ln x}{x^n+1}dx
\end{align}
where $\int_0^{\infty} \frac{x^{n-2}\ln x}{x^n+1}dx=\frac{\pi^2}{n^2}\csc\frac\pi n\cot\frac \pi n$
\begin{align}
&\int_0^{\infty} \frac{(1+x^{n-2})\ln \left(x^n+1\right)}{x^n+1}dx\\
=& \int_0^{\infty}\int_0^1 \frac{nt^{n-1}(1+x^{n-2})}{(x^n+1)(t^n+x^n)}dt\ dx\\
=& \int_0^1 \frac{2t^{n-1}-t^{n-2}-1}{t^n-1}\int_0^\infty\frac n{x^n+1}dx \ dt \\
=&\ \frac\pi2\csc\frac\pi n \int_0^1 \frac{2t^{n-1}-t^{n-2}-1}{t^n-1}dt\\
=& \ \frac{4\pi}n \csc\frac\pi n 
\bigg(\frac n4\ln2 - \sum_{k=1}^{[\frac n2]}\frac{\ln\csc\frac{k\pi}n}{\csc^2\frac{k\pi}n}\bigg)
\end{align}
Substitute above results into $I_n$ to obtain
$$I_n= \frac{4\pi}n \csc\frac\pi n 
\bigg(\frac n4\ln2 -\frac\pi8 \cot\frac\pi n- \sum_{k=1}^{[\frac n2]}\frac{\ln\csc\frac{k\pi}n}{\csc^2\frac{k\pi}n}\bigg)
$$
In particular, the close-form yields
\begin{align}
I_2&=\pi \ln2\\
I_3 &=\frac\pi {\sqrt3}\ln3-\frac{\pi^2}9\\
 I_4 &=\frac{3\pi}{2\sqrt2}\ln2-\frac{\pi^2}{4\sqrt2}\\
I_5 &= \frac{\pi\sqrt{\phi}}{\sqrt[4]5}\left(\frac12\ln5+\frac1{\sqrt5}\ln\phi\right)-\frac{\pi^2\phi^2}{5\sqrt5}\\
I_6&= \frac{2\pi}3\ln2+\frac\pi2\ln3 -\frac{\pi^2}{2\sqrt3}\\
I_7&= \frac{4\pi}7\csc\frac\pi7\bigg( \frac74\ln2-\frac\pi8\cot\frac\pi7 - \frac{\ln\csc\frac{\pi}7}{\csc^2\frac{\pi}7}- \frac{\ln\csc\frac{2\pi}7}{\csc^2\frac{2\pi}7}-\frac{\ln\csc\frac{3\pi}7}{\csc^2\frac{3\pi}7}\bigg)\\
I_8&= \pi\sqrt{1+\frac1{\sqrt2}}\left(\ln2+\frac1{2\sqrt2}\ln(\sqrt2+1)-\frac\pi8(\sqrt2+1)\right)\\
 I_9&= \frac{4\pi}9\csc\frac\pi9\bigg( \frac32\ln2+\frac38\ln3-\frac\pi8\cot\frac\pi9\\
&\hspace{3cm} - \frac{\ln\csc\frac{\pi}9}{\csc^2\frac{\pi}9}- \frac{\ln\csc\frac{2\pi}9}{\csc^2\frac{2\pi}9}-\frac{\ln\csc\frac{4\pi}9}{\csc^2\frac{4\pi}9}\bigg)\\
 I_{10}&= \frac{\pi \phi }{10}\bigg(\frac5{2}\ln5+4\ln2 +3\sqrt5\ln\phi-\pi\sqrt[4]5 \ \phi^{3/2}\bigg)\\
\end{align}
A: You can obtain the antiderivative. Let $x^n=t$ to face
$$I=\frac 1 n \int \frac{\log (t+1)}{t+1}\,t^{\frac{1}{n}-1}\,dt $$
$$I=n (t+1)^{\frac{1}{n}} \,
   _3F_2\left(-\frac{1}{n},-\frac{1}{n},-\frac{1}{n};1-\frac{1}{n},1
   -\frac{1}{n};\frac{1}{t+1}\right)+t^{\frac{1}{n}} \log (t+1)-$$
$$\frac{n t^{\frac{1}{n}+1} }{n+1}\,
   _2F_1\left(1,1+\frac{1}{n};2+\frac{1}{n};-t\right)
-(t+1)^{\frac{1}{n}} \log (t+1) \,
   _2F_1\left(-\frac{1}{n},-\frac{1}{n};\frac{n-1}{n};\frac{1}{t+1}\right)$$
A: For large $n$, a simple asymptotic behaviour of the integral can be deduced.
Let us first examine the structure of the integrand
$$y(x)=\frac{\ln \left(x^n+1\right)}{x^n+1} $$
Take the derivative of the function and set the result to zero to get the point $x=x_{m}$ where $y(x)$ reaches his maximum
$$1+(x_{m})^{n}=e$$
Putting this into $y(x)$ gives
$$y_{max}=y(x_{m})=\frac{\ln e}{e}=\frac{1}{e}$$
$y_{max}$ is independent of $n$
For integrals whose integrand, $y(x)$, has a sharp maximum a simple asymptotic formula exists.
$$I\approx \sqrt{2\pi\frac{y^{3}(x_{m})}{\left|y''(x_{m}) \right|}}$$
Here $y''(x_{m})$ is the second derivative of $y(x)$ at $x=x_{m}$.
I will skip elementary computations and write down final result
$$I_{n}\approx \frac{\sqrt{2\pi}}{n(e-1)^{1-\frac{1}{n}}}$$
Below is a few numerical examples to show approximation errors produced by the last formula
$n=10$,  the approximation error about $0.03$
$n=20$, the approximation error about $0.01$
$n=30$, the approximation error about $0.007$
The higher n is, the higher the accuracy of the last formula.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\color{#44f}{{\tt I}_{n}} & \equiv
\color{#44f}{\int_{0}^{\infty}{\ln\pars{x^{n} + 1} \over x^{n} + 1}\dd x} \sr{x^{n}\ \mapsto\ x}{=}
\int_{0}^{\infty}{\ln\pars{1 + x} \over 1 + x}\
{1 \over n}\, x^{1/n - 1}\,\,\dd x
\\[5mm] & =
\left.{1 \over n}\partiald{}{\nu}\int_{0}^{\infty}x^{\color{red}{1/n} - 1}\,\,
\pars{1 + x}^{\nu - 1}\,\dd x\right\vert_{\nu\ =\ 0}.
\\[5mm] & \mbox{Note that}\ \pars{1 + x}^{\nu - 1} =
\sum_{k = 0}^{\infty}{\nu - 1 \choose k}x^{k} =
\sum_{k = 0}^{\infty}\bracks{{k - \nu \choose k}\pars{-1}^{k}}x^{k}
\\[5mm] & =
\sum_{k = 0}^{\infty}{\Gamma\pars{k - \nu + 1} \over \Gamma\pars{1 - \nu}}{\pars{-x}^{k} \over k!}\quad \mbox{such that}
\\[5mm] \color{#44f}{{\tt I}_{n}} & \equiv
\color{#44f}{\int_{0}^{\infty}{\ln\pars{x^{n} + 1} \over x^{n} + 1}\dd x} =
{1 \over n}\partiald{}{\nu}\overbrace{\bracks{\Gamma\pars{\color{red}{1 \over n}}
{\Gamma\pars{-\color{red}{1/n} - \nu + 1} \over \Gamma\pars{1 - \nu}}}}^{\ds{Ramanujan's\ Master\ Theorem}}\hspace{-1mm}_{\substack{\\[2mm]\nu\ =\ 0}}
\\[5mm] & = \bbx{\color{#44f}{%
-\,{\pi\csc\pars{\pi/n}H_{-1/n}\,\, \over n}}} \\ &
\end{align}
