Help on how to show that $\int_{0}^{1}\left(2{x-1\over \ln^2{x}}-{x+1\over \ln{x}}\right)dx=3\ln{2}-2$ $$\int_{0}^{1}\left(2{x-1\over \ln^2{x}}-{x+1\over \ln{x}}\right)dx=3\ln{2}-2\tag1$$
Rewrite, so we can apply Frullani's formula on first part
$$\int_{0}^{1}\left(-{x+1\over \ln{x}}+{2\over \ln{x}}+{2(x-1)\over \ln^2{x}}\right)=3\ln{2}-2\tag2$$
$$-\ln{2}+\int_{0}^{1}\left(-{2\over \ln{x}}+{2(x-1)\over \ln^2{x}}\right)=3\ln{2}-2\tag3$$
$$\int_{0}^{1}\left(-{1\over \ln{x}}+{x-1\over \ln^2{x}}\right)=2\ln{2}-1\tag4$$
How do I continue from $(4)$?
 A: Through the substitution $x=e^{-t}$ the original integral equals:
$$ I=\int_{0}^{+\infty}\left(2\frac{e^{-t}-1}{t^2}+\frac{e^{-t}+1}{t}\right)e^{-t}\,dt \\=2\color{purple}{\int_{0}^{+\infty}\frac{e^{-t}-1+t}{t^2}\,e^{-t}\,dt}+\color{blue}{\int_{0}^{+\infty}\frac{e^{-t}-1}{t}\,e^{-t}\,dt}$$
where the blue integral is yet manageable through Frullani's theorem (leading to a $\log 2$), and the purple one becomes a similar integral by integration by parts, since:
$$ \frac{d}{dt}\left[(e^{-t}-1+t)e^{-t}\right]= 2(e^{-t}-e^{-2t})-te^{-t}.$$
By collecting pieces, $I=\color{red}{3\log 2-2}$ as wanted.
A: We can write the interal in $(4)$ as $$I=-\int_{0}^{1}\frac{1-x+\log\left(x\right)}{\log^{2}\left(x\right)}dx
 $$ now define $$I\left(\alpha\right)=-\int_{0}^{1}\frac{x^{\alpha}\left(1-x+\log\left(x\right)\right)}{\log^{2}\left(x\right)}dx,\,\alpha\geq0
 .
 $$ We have $$I''\left(\alpha\right)=-\int_{0}^{1}x^{\alpha}\left(1-x+\log\left(x\right)\right)dx=-\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\frac{1}{\left(\alpha+1\right)^{2}}
 $$ so integrating twice we have $$I\left(\alpha\right)=\left(2+\alpha\right)\log\left(\frac{2+\alpha}{1+\alpha}\right)+C\alpha+D
 $$ now since $$\lim_{\alpha\rightarrow\infty}I\left(\alpha\right)=0
 $$ and $$\lim_{\alpha\rightarrow\infty}\left(2+\alpha\right)\log\left(\frac{2+\alpha}{1+\alpha}\right)=1
 $$ we must have $C=0,\, D=-1
 $. So taking $\alpha=0
 $ we have 

$$I=I\left(0\right)=\color{red}{2\log\left(2\right)-1}$$ 

as wanted.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
&\color{#f00}{\int_{0}^{1}}\overbrace{\bracks{\color{#f00}{2\,{x - 1 \over \ln^2\pars{x}} - {x + 1 \over \ln\pars{x}}}}}
^{\ds{\int_{0}^{1}\pars{1 - 2t}x^{t}\,\dd t}}\color{#f00}{\,\dd x} =
\int_{0}^{1}\pars{1 - 2t}\int_{0}^{1}x^{t}\,\dd x\,\dd t =
\int_{0}^{1}{3 -2\pars{1 + t} \over t + 1}\,\dd t
\\[5mm] = &\
3\int_{0}^{1}{\dd t\over t + 1} - 2\int_{0}^{1}\dd t =
\color{#f00}{3\ln\pars{2} - 2}
\end{align}
