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

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 $$ 12\int_0^1\frac{\ln(1+x)}{x}dx=-12 \:\text{Li}_2(-1)=\pi^2. $$


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.

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Integration

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