A quicker method of proving $\int_{0}^{1}{6x(x-1)(x+2)\over (x+1)^3}\ln(x)dx=(\pi-3)(\pi+3)$ 
$$I=\int_{0}^{1}{6x(x-1)(x+2)\over (x+1)^3}\ln(x)dx=(\pi-3)(\pi+3)\tag1$$

$$I=\int_{0}^{1}\left(6-{12\over 1+x}-{6\over (1+x)^2}+{12\over (1+x)^3}\right)\ln(x)dx\tag2$$
Recall
$$\int_{0}^{1}{\ln(x)\over 1+x}dx=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{1}x^n\ln(x)dx=-\sum_{n=0}^{\infty}{(-1)^n\over (n+1)^2}=-{\pi^2\over 12}$$
$$\int_{0}^{1}{\ln(x)\over (1+x)^2}dx=\sum_{n=0}^{\infty}(-1)^n(n+1)\int_{0}^{1}x^n\ln(x)dx=-\sum_{n=0}^{\infty}{(-1)^n(n+1)\over (n+1)^2}=-\ln(2)$$
see answer of felix marin
Substitute into (2)$\rightarrow$ (3)
$$I=-6+\pi^2+6\ln(2)-12\cdot{1\over 4}[2\ln(2)+1]\tag3$$
$$I=\pi^2-9=(\pi-3)(\pi+3)\tag4$$
Anyone can prove I using an another approach? (Prefer quicker technique)
 A: Hint. One may integrate by parts.
$$
\begin{align}
&\int_{0}^{1}{6x(x-1)(x+2)\over (x+1)^3}\ln(x)dx
\\\\&=\left.  \left(6x+\frac{6x}{(1+x)^2}-12 \ln(1+x)\right)\ln x\right|_0^1-\int_0^1\left(6+\frac6{(1+x)^2}-12\frac{\ln(1+x)}{x}\right)dx
\\\\&=0-(9-\pi ^2)
\\\\&=(\pi-3)(\pi+3)
\end{align}
$$ where we have used the standard result
$$
12\int_0^1\frac{\ln(1+x)}{x}dx=-12 \:\text{Li}_2(-1)=\pi^2.
$$
A: An alternative approach. Since:
$$\forall x\in(0,1),\qquad \frac{x(x-1)(x+2)}{(x+1)^3}=\sum_{n\geq 1}(-1)^n (n^2+2n-1)x^n \tag{1}$$
we have:
$$\int_{0}^{1}\frac{x(x-1)(x+2)}{(x+1)^3}\,\log(x)\,dx = \sum_{n\geq 1}'(-1)^{n+1}\frac{n^2+2n-1}{(n+1)^2}=\frac{3}{2}-2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}\tag{2} $$
where $\sum'$ has to be intended à-la-Cesàro/Abel/Borel: $\sum_{n\geq 1}'a_n = \lim_{x\to 0^+}\sum_{n\geq 1}a_n e^{-nx}.$
In the RHS of $(2)$ we may easily recognize $\eta(2)=\frac{\zeta(2)}{2}=\frac{\pi^2}{12}$ and the claim easily follows.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\dd}{\mathrm{d}}
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 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{I} & =
\int_{0}^{1}{6x\pars{x - 1}\pars{x + 2} \over \pars{x + 1}^{3}}\ln\pars{x}
\,\dd x
\\[4mm] & =
12\int_{0}^{1}{\ln\pars{x} \over \pars{1 + x}^{3}}\,\dd x -
6\int_{0}^{1}{\ln\pars{x} \over \pars{1 + x}^{2}}\,\dd x -
12\int_{0}^{1}{\ln\pars{x} \over 1 + x}\,\dd x +
6\ \overbrace{\int_{0}^{1}\ln\pars{x}\,\dd x}^{\ds{-1}}
\end{align}

\begin{align}
\fbox{$\ds{\
\int_{0}^{1}{\ln\pars{x} \over a + x}\,\dd x\ }$} & =
-\int_{0}^{1}{\ln\pars{x} \over 1 - x/\pars{-a}}\,{\dd x \over -a} =
-\int_{0}^{-1/a}\,{\ln\pars{-ax} \over 1 - x}\,\dd x
\\[4mm] & =
-\int_{0}^{-1/a}{\ln\pars{1 - x} \over x}\,\dd x =
\fbox{$\ds{\ \Li{2}\pars{-\,{1 \over a}}\ }$}
\end{align}

$$
\left\lbrace\begin{array}{rcccl}
\ds{\int_{0}^{1}{\ln\pars{x} \over x + 1}} & \ds{=} &
\ds{\Li{2}\pars{-1}} & \ds{=} & \ds{-\,{\pi^{2} \over 12}}
\\[3mm]
\ds{\int_{0}^{1}{\ln\pars{x} \over \pars{x + 1}^{2}}} & \ds{=} &
\ds{\left.-\,\totald{\Li{2}\pars{-1/a}}{a}\right\vert_{\ a\ =\ 1}} & \ds{=} &
\ds{-\ln\pars{2}}
\\[3mm]
\ds{\int_{0}^{1}{\ln\pars{x} \over \pars{x + 1}^{3}}} & \ds{=} &
\ds{\left.\half\,\totald[2]{\Li{2}\pars{-1/a}}{a}\right\vert_{\ a\ =\ 1}}
& \ds{=} &
\ds{-\,{1 \over 4} - \half\,\ln\pars{2}}
\end{array}\right.
$$

\begin{align}
\color{#f00}{I} & =
\int_{0}^{1}{6x\pars{x - 1}\pars{x + 2} \over \pars{x + 1}^{3}}\ln\pars{x}
\,\dd x
\\[4mm] & =
12\bracks{-\,{1 \over 4} - \half\,\ln\pars{2}} -
6\bracks{-\ln\pars{2}} - 12\pars{-\,{\pi^{2} \over 12}} -6 =
\pi^{2} - 9 = \color{#f00}{\pars{\pi - 3}\pars{\pi + 3}}
\end{align}
