How prove this integral $\int_{0}^{1}\frac{x^m\ln{x}}{x-1}dx=\sum_{r=m+1}^{\infty}\frac{1}{r^2}$ prove that
$$\int_{0}^{1}\dfrac{x^m\ln{x}}{x-1}dx=\sum_{r=m+1}^{\infty}\dfrac{1}{r^2}$$
This true?
my idea
let
$$I(m)=\int_{0}^{1}\dfrac{x^m}{x-1}dx\Longrightarrow I'(m)=\int_{0}^{1}\dfrac{x^m\ln{x}}{x-1}dx$$
and we must find
$$I(m)=\int_{0}^{1}\dfrac{x^m}{x-1}dx=-\int_{0}^{1}\dfrac{(1-t)^m}{t}dt$$
then we can't prove it.
Thank you everyone can help, and This problem is from:How prove $\frac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{n}\frac{1}{t^2}<e$
 A: Here is how,
$$ \int_{0}^{1}\dfrac{x^m\ln{x}}{x-1}dx=-\sum_{k=0}^{\infty}\int_{0}^{1}x^{m+k}\ln(x)dx$$ 
$$= \sum_{k=0}^{\infty}\frac{1}{ (k+m)^2+2(k+m)+1 }$$
$$ = \sum_{k=m}^{\infty}\frac{1}{ k^2+2k+1 }$$
$$ = \sum_{k=m}^{\infty}\frac{1}{ (k+1)^2 }= \sum_{r=m+1}^{\infty}\frac{1}{ r^2 }=\psi'(m+1),$$
where $\psi(x)$ is the digamma function.
Note:
$$ \sum_{m+1}^{\infty}\frac{1}{r^2}=\psi'(m+1). $$
A: Hint: $\frac{1}{x-1} = -(1 + x + x^2 + \ldots)$.
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$\ds{\int_{0}^{1}{x^{m}\ln\pars{x} \over x - 1}\,\dd x
     =\sum_{r = m + 1}^{\infty}{1 \over r^{2}}}$

\begin{align}&\color{#66f}{\large\int_{0}^{1}{x^{m}\ln\pars{x} \over x - 1}\,\dd x}
=\lim_{\mu \to m}\partiald{}{\mu}\int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,\dd x
=\lim_{\mu \to m}\partiald{\Psi\pars{\mu + 1}}{\mu}=\Psi\,'\pars{m + 1}
\\[3mm]&=\partiald{}{z}\sum_{r = 1}^{\infty}\pars{{1 \over r} - {1 \over r + z}}
_{z\ =\ m }
=\sum_{r = 1}^{\infty}{1 \over \pars{r + m}^{2}}
=\color{#66f}{\large\sum_{r\ =\ m + 1}^{\infty}{1 \over r^{2}}}
\end{align}

See ${\bf 6.3.22}$ and
${\bf 6.3.16}$.
