Is that note of any use?

Let $t=\ln x$ $\begin{array}{c}\text{I} &= \int_{-\infty}^{0} \frac{t e^t}{1-e^{2t}} \text{d} t \\ &=\int_{0}^{\infty} \frac{t e^{-t}}{1-e^{-2t}} \text{d} t \\ &=\int_{0}^{\infty} t e^{-t} \left(\sum_{n=0}^{\infty} e^{-2nt} \right) \text{d} t \\ &=\int_{0}^{\infty} t \left(\sum_{n=0}^{\infty} e^{-(2n+1)t} \right) \text{d} t \\ &= \sum_{n=0}^{\infty} \left(\int_{0}^{\infty} te^{-(2n+1)t} \text{d} t \right) \\ &=-\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}\\ &=-\frac{\pi^2}{8} \end{array}$

$0<x<1\Rightarrow \frac { 1 }{ 1-x^{ 2 } } = \sum _{ k=0 }^{ \infty }{ { x }^{ 2k } } \color{#3D99F6} \text{ // geometric series} \\ I=\int _{ 0 }^{ 1 } \frac { \ln { ( } x) }{ 1-x^{ 2 } } dx=\int _{ 0 }^{ 1 } \sum _{ k=0 }^{ \infty }{ { x }^{ 2k }\cdot \ln { ( } x) } dx=\sum _{ k=0 }^{ \infty }{ { \int _{ 0 }^{ 1 } x }^{ 2k }\cdot \ln { ( } x) } dx \\ I=\sum _{ k=0 }^{ \infty }{ { \left[ \frac { { x }^{ 2k+1 } }{ 2k+1 } \ln { ( } x)|_{ 0 }^{ 1 } \right] -\frac { 1 }{ 2k+1 } \int _{ 0 }^{ 1 }{ { x }^{ 2k }dx } } } =-\sum _{ k=0 }^{ \infty }{ \frac { 1 }{ { \left( 2k+1 \right) }^{ 2 } } } \color{#3D99F6} \text{ // integration by part} \\ \color{#D61F06} I \text{ is only the odd squares} \\ \text{so }I=-\left[ \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 } } } -\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { \left( 2k \right) }^{ 2 } } } \right] =-\left[ \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 } } } -\frac { 1 }{ 4 } \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 } } } \right] \\ I=-\left[ \frac { { \pi }^{ 2 } }{ 6 } -\frac { 1 }{ 4 } \cdot \frac { { \pi }^{ 2 } }{ 6 } \right] =-\frac { { \pi }^{ 2 } }{ 8 }$