Difficult Integral Involving the $\ln$ function Please help me solve this integral!
I have tried multiple different procedures for integration by parts, as well as substitution and have not come up with anything.
$$\int\frac{\ln x}{(\ln x+1)^2}dx$$
Thank you
 A: Setting $\ln x=y\implies x=e^y$
$$\int\frac{\ln x}{(\ln x+1)^2}dx=\int\frac y{(y+1)^2}e^y dy$$
$$=\int\left(\frac{y+1-1}{(y+1)^2}\right) e^ydy=\int e^y\left(\frac1{y+1}-\frac1{(y+1)^2}\right)$$
If $\displaystyle f(y)=\frac1{y+1}, f'(y)=?$
Now,
$$\int e^y\left[f(y)+f'(y)\right]dy=f(y)e^ydy+f'(y)e^y=d(e^yf(y))$$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\mbox{Note that}\quad\int{\ln\pars{x} \over \bracks{\ln\pars{x} + 1}^{2}}\,\dd x
=\left.-\,\partiald{}{\mu}\int{\dd x \over \mu\ln\pars{x} + 1}
\right\vert_{\mu\ =\ 1}
\end{align}

\begin{align}
&\int{\dd x \over \mu\ln\pars{x} + 1}={1 \over \mu}
\int{\dd x \over \ln\pars{x} + 1/\mu}
={\expo{-1/\mu} \over \mu}\
\overbrace{\int{\expo{1/\mu}\dd x \over \ln\pars{x\expo{1/\mu}}}}
^{\ds{\mbox{Set}\ t\equiv\expo{1/\mu}x}}\ =\ 
{\expo{-1/\mu} \over \mu}\int{\dd t \over\ln\pars{t}}
\\[3mm]&={\expo{-1/\mu} \over \mu}\bracks{{\rm li}\pars{t}
+\pars{~\mbox{a constant}~}}
={\expo{-1/\mu} \over \mu}\bracks{{\rm li}\pars{\expo{1/\mu}x}
+\pars{~\mbox{a constant}~}}
\end{align}

$\ds{{\rm li}\pars{x}}$ is the
Logarithmic Integral Function.

\begin{align}
&\color{#66f}{\large\int{\ln\pars{x} \over \bracks{\ln\pars{x} + 1}^{2}}\,\dd x}
=\left.-\,\partiald{}{\mu}\braces{
{\expo{-1/\mu} \over \mu}\bracks{{\rm li}\pars{\expo{1/\mu}x}
+\pars{~\mbox{a constant}~}}}\right\vert_{\mu\ =\ 1}
\\[3mm]&=\color{#66f}{\large{x \over \ln\pars{x} + 1}} + \pars{~\mbox{a constant}~}
\end{align}

A: Integration by parts can be made to work
$$\int \frac{\log x}{(1 + \log x)^2} = \int x \log x \frac{1}{x(1+ \log x)^2} = \int x\log x \left(\frac{-1}{\log x + 1}\right)'  =$$
$$\frac{-x\log x}{\log x + 1} + \int (x \log x)' \frac{1}{\log x + 1} = \frac{-x\log x}{\log x + 1} + \int (1 + \log x) \frac{1}{\log x + 1} = $$
$$\frac{-x\log x}{\log x + 1} + x + C = \frac{x}{\log x + 1} + C$$
