Find the integral $\int_0^1 \frac{x^2 - 1}{ \ln(x)} dx$ I tried substituting $\ln (x)$ as $t$, but it led to no standard integral for me to move further.
Substituting $\ln (x)$ as $-t$ gives
$$
\int_0^\infty\frac{e^{-t}-e^{-3t}}{t}\,\mathrm{d}t
$$

Expanding on Robert Z's answer,
$$
\begin{align}
\int_0^1\frac{x^2-1}{\log(x)}\,\mathrm{d}x
&=-\int_0^\infty\frac{e^{-2x}-1}{x}\,e^{-x}\,\mathrm{d}x\\
&=\int_0^\infty\frac{1-e^{-2x}}{x}\,e^{-x}\,\mathrm{d}x\\
&=\int_0^\infty\frac{e^{-x}-e^{-3x}}{x}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-x}-e^{-3x}}{x}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\left(\int_\epsilon^\infty\frac{e^{-x}}{x}\,\mathrm{d}x-\int_\epsilon^\infty\frac{e^{-3x}}{x}\,\mathrm{d}x\right)\\
&=\lim_{\epsilon\to0^+}\left(\int_\epsilon^\infty\frac{e^{-x}}{x}\,\mathrm{d}x-\int_{3\epsilon}^\infty\frac{e^{-x}}{x}\,\mathrm{d}x\right)\\
&=\lim_{\epsilon\to0^+}\int_\epsilon^{3\epsilon}\frac{e^{-x}}{x}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\int_1^3\frac{e^{-\epsilon x}}{x}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\int_1^3\left(\frac1x+\frac{o(\epsilon)}x\right)\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\left(\int_1^3\frac1x\,\mathrm{d}x+o(\epsilon)\right)\\
&=\int_1^3\frac1x\,\mathrm{d}x\\[9pt]
&=\log(3)
\end{align}
$$
