Find the integral $\int_0^1 \frac{x^2 - 1}{ \ln(x)} dx$

In general, for $a>-1$, by letting $t=-\ln(x)$, we obtain $$\int_0^1 \frac{x^a-1}{\ln x} dx=\int_0^{+\infty} \frac{e^{-t}-e^{-(a+1)t}}{t} dt=\ln(a+1)$$ where in the last step we used a known result about Frullani's integral with $f(t)=e^{-t}$.

Hint: for $a>0$, write $$f(a) = \int_0^1 \frac{x^a-1}{\ln x} dx$$ Then we can differentiate with respect to $a$ to give $f'(a) = 1/(a+1)$. Also note that $$\lim_{a\to 0^+} f(a) = 0$$ This can help you to find the integration constant.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{x^{2} - 1 \over \ln\pars{x}}\,\dd x & = \int_{0}^{1}\pars{x + 1}\ \overbrace{{x - 1 \over \ln\pars{x}}}^{\ds{= \int_{0}^{1}x^{t}\,\dd t}}\ \,\dd x = \int_{0}^{1}\int_{0}^{1}\pars{x^{t + 1} + x^{t}}\,\dd x\,\dd t \\[5mm] & = \int_{0}^{1}\pars{{1 \over t + 2} + {1 \over t + 1}}\,\dd t = \bracks{\ln\pars{3} - \ln\pars{2}} + \ln\pars{2} \\[5mm] & = \bbx{\ln\pars{3}} \approx 1.099 \end{align}