[math(\displaystyle \begin{aligned}
{\sf Let}: I_n &= \int \frac{\ln^nu}{u^2} \,{\rm d}u \\
\Rightarrow I_0 &= \int \frac1{u^2} \,{\rm d}u = -\frac1u + {\sf const}. \\
\therefore I_n &= \int \frac{\ln^nu}{u^2} \,{\rm d}u = \int \ln^nu \cdot \frac1{u^2} \,{\rm d}u \\
&= \ln^nu \cdot \!\left( -\frac1u \right) -\int n\ln^{n-1}u \cdot \frac1u \cdot \!\left( -\frac1u \right) \!{\rm d}u \\
&= -\frac{\ln^nu}u +n\int \frac{\ln^{n-1}u}{u^2} \,{\rm d}u \\
&= -\frac{\ln^nu}u +nI_{n-1} \\
&= -\frac{\ln^nu}u -\frac{n\ln^{n-1}u}u +n(n-1)I_{n-2} \\
&= \cdots \\
&= -\frac{\ln^nu}u -\frac{n\ln^{n-1}u}u -\cdots -\frac{n(n-1)\cdots3\cdot2\cdot\ln u}u +n(n-1)\cdots2\cdot1\cdot I_0 \\
&= -\frac{\ln^nu}u -\frac{n\ln^{n-1}u}u -\cdots -\frac{n(n-1)\cdots3\cdot2\cdot\ln u}u -\frac{n!}u +{\sf const}. \\
&= -\sum_{i=0}^n \frac{n^{\underline i} \ln^{n-i}u}u +{\sf const}. \\
\therefore \int \frac{\ln^nu}{u^2} \,{\rm d}u &= -\sum_{i=0}^n \frac{n^{\underline i} \ln^{n-i}u}u +{\sf const}.
\end{aligned} )]
[math(\displaystyle \begin{aligned}
\int_k^{k+1} \frac{k\ln^nu}{u^2} \,{\rm d}u &= \!\Biggl[ -\sum_{i=0}^n \frac{kn^{\underline i} \ln^{n-i}u}u \Biggr]_k^{k+1} \\
&= -\sum_{i=0}^n \frac{kn^{\underline i} \ln^{n-i} (k+1)}{k+1} +\sum_{i=0}^n n^{\underline i} \ln^{n-i}k \\
&= -\frac k{k+1} \sum_{i=0}^n n^{\underline i} \ln^{n-i} (k+1) +\sum_{i=0}^n n^{\underline i} \ln^{n-i}k \\
&= -\biggl( 1-\frac1{k+1} \biggr) \!\sum_{i=0}^n n^{\underline i} \ln^{n-i} (k+1) +\sum_{i=0}^n n^{\underline i} \ln^{n-i}k \\
&= \frac1{k+1} \Biggl( \sum_{i=0}^n n^{\underline i} \ln^{n-i} (k+1) \Biggr) \!+\!\Biggl( \sum_{i=0}^n n^{\underline i} \ln^{n-i}k \Biggr) \!-\!\Biggl( \sum_{i=0}^n n^{\underline i} \ln^{n-i} (k+1) \Biggr) \\
&= \frac1{k+1} \Biggl( {\color{DeepSkyBlue} \ln^n (k+1) } +\sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} n! } \Biggr) \\
&\quad +\!\Biggl( \sum_{i=0}^{n-2} n^{\underline i} \ln^{n-i}k +{\color{limegreen} n^{\underline{n-1}} \ln k } +n! \Biggr) \!-\!\Biggl( \sum_{i=0}^{n-2} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} n^{\underline{n-1}} \ln (k+1) } +n! \Biggr) \\
&= {\color{DeepSkyBlue} \frac{\ln^n (k+1)}{k+1} } +\frac1{k+1} \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} \frac{n!}{k+1} } \\
&\quad +\sum_{i=0}^{n-2} n^{\underline i} (\ln^{n-i}k -\ln^{n-i} (k+1)) +{\color{limegreen} n^{\underline{n-1}} (\ln k -\ln (k+1)) } \\
&= {\color{DeepSkyBlue} \frac{\ln^n (k+1)}{k+1} } +\frac1{k+1} \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} \frac{n!}{k+1} } \\
&\quad +\sum_{i=1}^{n-1} n^{\underline{i-1}} (\ln^{n-i+1}k -\ln^{n-i+1} (k+1)) +{\color{limegreen} n^{\underline{n-1}} (\ln k -\ln (k+1)) }
\end{aligned} )]
[math(\displaystyle \begin{aligned}
\sum_{k=1}^{m-1} &\int_k^{k+1} \frac{k\ln^nu}{u^2} \,{\rm d}u {\color{DeepSkyBlue} \,-\,\frac{\ln^{n+1}m}{n+1} } \\
&= {\color{DeepSkyBlue} -\frac{\ln^{n+1}m}{n+1} } +\sum_{k=1}^{m-1} \Biggl( {\color{DeepSkyBlue} \frac{\ln^n (k+1)}{k+1} } +\frac1{k+1} \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} \frac{n!}{k+1} } \\
&\quad +\sum_{i=1}^{n-1} n^{\underline{i-1}} (\ln^{n-i+1}k -\ln^{n-i+1} (k+1)) +{\color{limegreen} n^{\underline{n-1}} (\ln k -\ln (k+1)) } \Biggr) \\
&= {\color{DeepSkyBlue} -\frac{\ln^{n+1}m}{n+1} +\sum_{k=1}^{m-1} \frac{\ln^n (k+1)}{k+1} } +\sum_{k=1}^{m-1} \frac1{k+1} \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i} (k+1) +{\color{limegreen} \sum_{k=1}^{m-1} \frac{n!}{k+1} } \\
&\quad +\sum_{k=1}^{m-1} \sum_{i=1}^{n-1} n^{\underline{i-1}} (\ln^{n+1-i}k -\ln^{n+1-i} (k+1)) +{\color{limegreen} \sum_{k=1}^{m-1} n^{\underline{n-1}} (\ln k -\ln (k+1)) } \\
&= {\color{DeepSkyBlue} -\frac{\ln^{n+1}m}{n+1} +\sum_{k=2}^m \frac{\ln^n k}k } +\sum_{k=2}^m \frac1k \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i}k +{\color{limegreen} \sum_{k=2}^m \frac{n!}k } \\
&\quad +\sum_{i=1}^{n-1} n^{\underline{i-1}} \sum_{k=1}^{m-1} (\ln^{n+1-i}k -\ln^{n+1-i} (k+1)) {\color{limegreen} \,-\,n^{\underline{n-1}} \ln m } \\
&= {\color{DeepSkyBlue} \sum_{k=2}^m \frac{\ln^n k}k -\frac{\ln^{n+1}m}{n+1} } +\sum_{k=2}^m \frac1k \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i}k +{\color{limegreen} \sum_{k=2}^m \frac{n!}k } +\sum_{i=1}^{n-1} n^{\underline{i-1}} (-\ln^{n+1-i}m) {\color{limegreen} \,-\,n^{\underline{n-1}} \ln m } \\
&= {\color{DeepSkyBlue} \sum_{k=1}^m \frac{\ln^n k}k -\frac{\ln^{n+1}m}{n+1} } +\sum_{k=1}^m \frac1k \sum_{i=1}^{n-1} n^{\underline i} \ln^{n-i}k +{\color{limegreen} \sum_{k=1}^m \frac{n!}k -n! } -\sum_{i=1}^{n-1} n^{\underline{i-1}} \ln^{n+1-i}m {\color{limegreen} \,-\,n^{\underline{n-1}} \ln m } \\
&= {\color{DeepSkyBlue} \sum_{k=1}^m \frac{\ln^n k}k -\frac{\ln^{n+1}m}{n+1} } +\sum_{i=1}^{n-1} \sum_{k=1}^m \frac{n^{\underline i} \ln^{n-i}k}k +{\color{limegreen} \sum_{k=1}^m \frac{n!}k -n! } -\sum_{i=1}^{n-1} n^{\underline{i-1}} \ln^{n-i+1}m {\color{limegreen} \,-\,n^{\underline{n-1}} \ln m } \\
&= {\color{DeepSkyBlue} \!\Biggl( \sum_{k=1}^m \frac{\ln^n k}k -\frac{\ln^{n+1}m}{n+1} \Biggr) } \!+\sum_{i=1}^{n-1} \Biggl( \sum_{k=1}^m \frac{n^{\underline i} \ln^{n-i}k}k -n^{\underline{i-1}} \ln^{n-i+1}m \Biggr) \!+{\color{limegreen} \!\Biggl( \sum_{k=1}^m \frac{n!}k -n^{\underline{n-1}} \ln m \Biggr) \!-n! } \\
&\qquad {\sf Note\ 1}: n^{\underline{i-1}} = \frac{n!}{(n-(i-1))!} = \frac{n!}{(n-i+1)!} = \frac{n!}{(n-i)!}\cdot\frac1{n-i+1} = \frac{n^{\underline i}}{n-i+1} \\
&\qquad {\sf Note\ 2}: n^{\underline{n-1}} = \frac{n!}{1!} = \frac{n!}{0!\cdot1} = \frac{n^{\underline n}}1 \\
&= \!\Biggl( \sum_{k=1}^m \frac{\ln^n k}k -\frac{\ln^{n+1}m}{n+1} \Biggr) \!+\sum_{i=1}^{n-1} \Biggl( \sum_{k=1}^m \frac{n^{\underline i} \ln^{n-i}k}k -\frac{n^{\underline i} \ln^{n-i+1}m}{n-i+1} \Biggr) \!+\!\Biggl( \sum_{k=1}^m \frac{n!}k -\frac{n^{\underline n} \ln m}1 \Biggr) \!-n! \\
&= \sum_{i=0}^n \Biggl( \sum_{k=1}^m \frac{n^{\underline i} \ln^{n-i}k}k -\frac{n^{\underline i} \ln^{n-i+1}m}{n-i+1} \Biggr) \!-n! \\
&= \sum_{i=0}^n n^{\underline i} \Biggl( \sum_{k=1}^m \frac{\ln^{n-i}k}k -\frac{\ln^{n-i+1}m}{n-i+1} \Biggr) \!-n!
\end{aligned} )]
[math(\displaystyle \begin{aligned}
\therefore \int_0^1 \!\left\{ \frac1x \right\} \ln^nx \,{\rm d}x &= \int_\infty^1 \{u\} \ln^n u^{-1} \biggl( \!-\frac{{\rm d}u}{u^2} \biggr) \\
&= (-1)^{n} \int_1^\infty \frac{u-\!\left\lfloor u \right\rfloor}{u^2} \ln^nu \,{\rm d}u \\
&= (-1)^{n+1} \int_1^\infty \frac{\left\lfloor u \right\rfloor \!-u}{u^2} \ln^nu \,{\rm d}u \\
&= (-1)^{n+1} \lim_{m\to\infty} \int_1^m \biggl( \frac{\left\lfloor u \right\rfloor \!\ln^nu}{u^2} -\frac{\ln^nu}u \biggr) {\rm d}u \\
&= (-1)^{n+1} \lim_{m\to\infty} \Biggl( \sum_{k=1}^{m-1} \int_k^{k+1} \frac{\left\lfloor u \right\rfloor \!\ln^nu}{u^2} \,{\rm d}u -\int_1^m \frac{\ln^nu}u \,{\rm d}u \Biggr) \\
&= (-1)^{n+1} \lim_{m\to\infty} \Biggl( \sum_{k=1}^{m-1} \int_k^{k+1} \frac{k\ln^nu}{u^2} \,{\rm d}u -\biggl[ \frac{\ln^{n+1} u}{n+1} \biggr]_1^m \Biggr) \\
&= (-1)^{n+1} \lim_{m\to\infty} \Biggl( \sum_{k=1}^{m-1} \int_k^{k+1} \frac{k\ln^nu}{u^2} \,{\rm d}u -\frac{\ln^{n+1} m}{n+1} \Biggr) \\
&= (-1)^{n+1} \lim_{m\to\infty} \Biggl( \sum_{i=0}^n n^{\underline i} \Biggl( \sum_{k=1}^m \frac{\ln^{n-i}k}k -\frac{\ln^{n-i+1}m}{n-i+1} \Biggr) \!-n! \Biggr) \\
&= (-1)^{n+1} \Biggl( \sum_{i=0}^n n^{\underline i} \lim_{m\to\infty} \Biggl( \sum_{k=1}^m \frac{\ln^{n-i}k}k -\frac{\ln^{n-i+1}m}{n-i+1} \Biggr) \!-n! \Biggr) \\
&= (-1)^{n+1} \Biggl( \sum_{i=0}^n n^{\underline i} \gamma_{n-i} -n! \Biggr) \\
\end{aligned} )]