Find $\int_0^1 \mathrm{\frac{x-1}{ln(x)}}\,\mathrm{d}x$ Find $\int_0^1 \mathrm{\frac{x-1}{ln(x)}}\,\mathrm{d}x$
I tryed this:
$\int_0^1 \mathrm{\frac{x-1}{ln(x)}} = \int_0^1 \mathrm{\frac{x}{ln(x)}} - \int_0^1 \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x$
To $\int_0^1 \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x$
Let $t=lnx $ then $\frac{dt}{dx}=\frac{1}{x}$ and $dx=e^tdt$
$\int \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x \equiv \int \mathrm{\frac{e^t}{t}}$$\mathrm{d}t$ and well
$e^t=\sum _{n=0}^\infty \frac{t^n}{n!}$ 
$\frac{e^t}{t}=\sum_{n=0}^\infty \frac{t^{n-1}}{n!}$ 
$\int \mathrm{\frac{e^t}{t}}\,\mathrm{d}t=\int \mathrm \sum_{n=0}^\infty{\frac{t^{n-1}}{n!}}\,\mathrm{d}t=\sum_{n=0}^\infty\frac{t^n}{n*(n)!}$
How can I solve $\int_0^1 \mathrm{\frac{x}{ln(x)}}$ ?
Thanks for your help :) have a nice day
 A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\int_{0}^{1}{x - 1 \over \ln\pars{x}}\,\dd x: {\large ?}.\quad}$
Whith the change of variables $x \equiv \expo{-z}:$
\begin{align}
\color{#0000ff}{\large\int_{0}^{1}{x - 1 \over \ln\pars{x}}\,\dd x}
&=
\int_{\infty}^{0}{\expo{-z} - 1 \over -z}\,\pars{-\expo{-z}\,\dd z}
=
\int^{\infty}_{0}{\expo{-z} - \expo{-2z}\over z}\,\dd z
\\[3mm]&=
-\int^{\infty}_{0}\ln\pars{z}\pars{-\expo{-z} +2 \expo{-2z}}\,\dd z
=
\int^{\infty}_{0}\ln\pars{z}\expo{-z}\,\dd z
-
\int^{\infty}_{0}\ln\pars{z \over 2}\expo{-z}\,\dd z
\\[3mm]&=
\int^{\infty}_{0}\ln\pars{z}\expo{-z}\,\dd z
-
\int^{\infty}_{0}\bracks{\ln\pars{z} - \ln\pars{2}}\expo{-z}\,\dd z
=
\ln\pars{2}\overbrace{\int^{\infty}_{0}\expo{-z}\,\dd z}^{=\ 1}
\\[3mm]&=
\color{#0000ff}{\large\ln\pars{2}}
\end{align}
