Showing that $\int_{0}^{1}{x(x-1)(x+2)\over (x+1)^3}\cdot{1\over \ln(x)}dx={\ln{\pi\over 2}-{7\zeta(3)\over 4\pi^2}}$ Showing that

$$\int_{0}^{1}{x(x-1)(x+2)\over (x+1)^3}\cdot{1\over \ln(x)}dx=\color{brown}{\ln{\pi\over 2}-{7\zeta(3)\over 4\pi^2}}$$

Applying substitution
$u=\ln(x)\rightarrow du={1\over x}dx$ and $x=e^u$
$x=1\rightarrow u=0$ and $x=0\rightarrow u=-\infty$
Then
$$I=\int_{-\infty}^{0}{e^{2u}(e^u-1)(e^u+2)\over u(e^u+1)^3}du$$
$${e^{2u}(e^u-1)(e^u+2)\over u(e^u+1)^3}=e^u-2+{e^{2u}+5e^u+2\over (e^u+1)^3}$$
Substitute back in
$$I=\int_{-\infty}^{0}{e^u-2\over u}+{e^{2u}+5e^u+2\over u(e^u+1)^3}du$$
$e^{2u}+5e^u+2=A(e^u+1)+Bu(e^u+1)^2+Cu(e^u+1)+Du$
$u=0\rightarrow A=1$
I think it is impossible to find the values of B,C and D.
This method is not working. Can anyone give me a hint on another method?

$$\int_{0}^{1}\frac{1-x}{(1+x)\log x}\,dx =\log\frac{\pi}{2}\tag{1}$$
is a straightforward consequence of Frullani's theorem and Wallis' product. 
So it is enough to compute:
$$\begin{eqnarray*} \int_{0}^{1}\frac{1-x}{(1+x)^3\log x}\,dx &=& \int_{0}^{+\infty}\frac{e^{-x}-e^{-2x}}{x(1+e^{-x})^3}\,dx\\&=&-7\cdot\zeta'(-2)\tag{2}\end{eqnarray*}$$
always by Frullani's theorem. The reflection formula for the $\zeta$ function finishes the job.
