Is there other method to evaluate $\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x, \textrm{ where }n\in N?$ Letting $x\mapsto \frac{1}{x}$ transforms the integral into
$\displaystyle I=\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x=-\int_0^1 \frac{x^{2 n-1} \ln x}{x^n+1} d x \tag*{} $
Splitting the integrand into two pieces like
$\displaystyle I=- \underbrace{\int_0^1 x^{n-1} \ln x d x}_{J} + \underbrace{\int_0^1 \frac{x^{n-1} \ln x}{x^n+1} d x}_{K} \tag*{} $

For the integral $J,$ letting $z=-n\ln x$ transforms $J$ into
$\displaystyle J= -\frac{1}{n^2} \int_0^{\infty} z e^{-z} d z =-\frac{1}{n^2}\tag*{} $

For integral $K$, using the series for $|x|<1,$
$\displaystyle \ln (1+x)=\sum_{k=0}^{\infty} \frac{(-1)^k}{k+1} x^{k+1},\tag*{} $
we have
$\displaystyle \begin{aligned}K& =-\frac{1}{n} \sum_{k=0}^{\infty} \frac{(-1)^k}{k+1} \int_0^1 x^{n(k+1)-1} d x \\& =-\frac{1}{n^2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2} \\& =-\frac{1}{n^2}\left[\sum_{k=1}^{\infty} \frac{1}{k^2}-2 \sum_{k=1}^{\infty} \frac{1}{(2 k)^2}\right] \\& =-\frac{1}{2 n^2} \cdot \frac{\pi^2}{6} \\& =-\frac{\pi^2}{12 n^2}\end{aligned}\tag*{} $

Putting them back yields
$\displaystyle \boxed{ I=\frac{1}{12 n^2}\left(12-\pi^2\right)}\tag*{} $

Is there alternative method?  Comments and alternative methods are highly appreciated.
 A: Substitute $t=\frac1{x^n}$
$$\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x
=-\frac1{n^2}\int_0^1 \frac{t\ln t}{1+t}dt
= \frac{1}{n^2}\left(1-\frac{\pi^2}{12}\right)
$$
A: Making it more general $$I=\int \frac{\log( x)}{x^{m}\left(1+x^n\right)}\, d x$$
$$x=t^{\frac{1}{n}} \quad \implies \quad I=\frac 1{n^2}\int \frac {\log(t)}{t^a\,(1+t)}\,dt \qquad \text{with} \qquad a=\frac{m+n-1}{n}$$ The antiderivative express in terms of two hypergeometric functions.
$$J=\int_1^\infty \frac{\log( x)}{x^{m}\left(1+x^n\right)}\, d x$$ $$\color{blue}{J=\frac 1{4n^2}\left(\psi ^{(1)}\left(\frac{m+n-1}{2 n}\right)-\psi ^{(1)}\left(\frac{m+2
   n-1}{2 n}\right)\right)}$$
Edit
If we make $m=n+k$ and consider large values of $n$
$$\psi ^{(1)}\left(\frac{2n+k-1}{2 n}\right)-\psi ^{(1)}\left(\frac{3n+k-1}{2 n}\right)=\left(4-\frac{\pi ^2}{3}\right)+$$ $$\frac{2 (k-1) (3 \zeta (3)-4)}{n}-\frac{ (k-1)^2\left(7
   \pi ^4-720\right)}{60 n^2}+O\left(\frac{1}{n^3}\right)$$
A: Thanks to @Quanto’s short solution and @Claude’s generalisation using hypergeometric function. I am going to use Feynman’s Integration technique to prove the generalisation
$$
\boxed{I=\int_1^{\infty} \frac{\ln x}{x^m\left(1+x^n\right)} d x=\frac{1}{4 n^2}\left[\zeta\left(2, \frac{m+n-1}{2 n}\right)-\zeta\left(2, \frac{m+2 n-1}{2 n}\right)\right]}
$$
Letting $x=t^{-\frac{1}{n}}$ transforms the integral into
$$
I=-\frac{1}{n^2} \int_0^1 \frac{t^{\frac{m-1}{n}} \ln t}{1+t} d t= -\frac{1}{n^2}\left.\frac{\partial}{\partial b} J(b)\right|_{b=\frac{m-1}{n}},
$$
where $J(b)=\int_0^1 \frac{t^b}{1+t} d t.$
Using power series, we have
$$
J(b) =\int_0^1 \frac{t^b}{1+t} d t=\sum_{k=0}^{\infty}(-1)^k \int_0^1 t^{k+k} d t=\sum_{k=0}^{\infty} \frac{(-1)^k}{b+k+1}
$$
Differentiating both sides w.r.t. $b$ and putting $b=\frac{m-1}{n}$ gives
$$
\begin{aligned}
\int_1^{\infty} \frac{\ln x}{x^m\left(1+x^n\right)} d x & = \frac{1}{n^2}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{\left(k+\frac{m+n-1}{n}\right)^2} \\
& =\frac{1}{4 n^2}\left[\zeta\left(2, \frac{m+n-1}{2 n}\right)-\zeta\left(2, \frac{m+2 n-1}{2 n}\right)\right],
\end{aligned}
$$
which matches exactly @Cluade’s answer
$$\color{blue}{J=\frac 1{4n^2}\left(\psi ^{(1)}\left(\frac{m+n-1}{2 n}\right)-\psi ^{(1)}\left(\frac{m+2
   n-1}{2 n}\right)\right)}$$
