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.
Best Answer
Substitute $t=\frac1{x^n}$ $$\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x =-\frac1{n^2}\int_0^1 \frac{t\ln t}{1+t}dt = \frac{1}{n^2}\left(1-\frac{\pi^2}{12}\right) $$