# 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

• Can you give us an idea of where you got stuck in your attempts? – Cameron Williams Feb 23 '14 at 19:02
• Everywhere haha. I mean i didn't know how to properly split up the integral into the two parts – Chrysanthemum Feb 23 '14 at 19:03

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, $$\int e^y\left[f(y)+f'(y)\right]dy=f(y)e^ydy+f'(y)e^y=d(e^yf(y))$$

• Thank you very much! So is there any particular recipe or method to solving these types of questions? What are the things that are important to recognize? – Chrysanthemum Feb 23 '14 at 19:12
• @user130996, the replacement of $ln x$ can be a good point to start. Then, taking out $e^y$ as common factor – lab bhattacharjee Feb 23 '14 at 19:14
• that was some tite methodology dude – enthdegree Feb 23 '14 at 19:31

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}

Integration by parts can be made to work

$$\int \frac{\log x}{(1 + \log x)^2} = \int x \log x \frac{1}{x(1+ \log x)^2} = \int x\log x \left(\frac{-1}{\log x + 1}\right)' =$$

$$\frac{-x\log x}{\log x + 1} + \int (x \log x)' \frac{1}{\log x + 1} = \frac{-x\log x}{\log x + 1} + \int (1 + \log x) \frac{1}{\log x + 1} =$$ $$\frac{-x\log x}{\log x + 1} + x + C = \frac{x}{\log x + 1} + C$$