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
$$
\int_0^\infty\frac{e^{-t}-e^{-3t}}{t}\,\mathrm{d}t
$$
 A: 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 $f(t)=e^{-t}$.
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\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}
A: Expanding on Robert Z's answer,
$$
\begin{align}
\int_0^1\frac{x^2-1}{\log(x)}\,\mathrm{d}x
&=-\int_0^\infty\frac{e^{-2x}-1}{x}\,e^{-x}\,\mathrm{d}x\\
&=\int_0^\infty\frac{1-e^{-2x}}{x}\,e^{-x}\,\mathrm{d}x\\
&=\int_0^\infty\frac{e^{-x}-e^{-3x}}{x}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-x}-e^{-3x}}{x}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\left(\int_\epsilon^\infty\frac{e^{-x}}{x}\,\mathrm{d}x-\int_\epsilon^\infty\frac{e^{-3x}}{x}\,\mathrm{d}x\right)\\
&=\lim_{\epsilon\to0^+}\left(\int_\epsilon^\infty\frac{e^{-x}}{x}\,\mathrm{d}x-\int_{3\epsilon}^\infty\frac{e^{-x}}{x}\,\mathrm{d}x\right)\\
&=\lim_{\epsilon\to0^+}\int_\epsilon^{3\epsilon}\frac{e^{-x}}{x}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\int_1^3\frac{e^{-\epsilon x}}{x}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\int_1^3\left(\frac1x+\frac{o(\epsilon)}x\right)\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\left(\int_1^3\frac1x\,\mathrm{d}x+o(\epsilon)\right)\\
&=\int_1^3\frac1x\,\mathrm{d}x\\[9pt]
&=\log(3)
\end{align}
$$
A: The antiderivative is not elementary: $F(x) = \text{Ei}(1,  -\ln(x)) - \text{Ei}(1,-3 \ln(x))$, where $$\text{Ei}(1,z) = \int_1^\infty e^{-tz} t^{-1}\; dt $$
As $x \to 0+$, $-\ln(x) \to +\infty$ and both $\text{Ei}(1,  -\ln(x))$ and $\text{Ei}(1,-3 \ln(x))$ go to $0$.  As $x \to 1-$, both
$\text{Ei}(1,-\ln(x))$ and $\text{Ei}(1,-3 \ln(x)) \to \infty$, but 
$$\text{Ei}(1, 1/s) \sim \ln(s) -\gamma + O(1/s)$$
resulting in $\text{Ei}(1,-\ln(x)) - \text{Ei}(1,-3 \ln(x)) = \ln(3) + O(\ln(x))$.
Thus the answer is $\ln(3)$.
