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How to integrate- $$\int \frac{1}{x\ln x+x}dx$$ ? I am trying this question by substituting $ln(x)$ $=$ $u$. So, $(1/x)dx$ $=$ $du$. Also $x$ will become $e^{u}$. So, finally the expression will become $$\int [1/(u+e^{u})]du$$. But I can't approach further. Please help me out to evaluate the integral.

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    $\begingroup$ Hi, welcome to Math SE. Wolfram Alpha finds no answer in terms of standard functions. $\endgroup$
    – J.G.
    Commented Oct 30, 2023 at 13:43
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    $\begingroup$ Hint: $x\ln(x)+x = x(\ln(x)+1)$. And you made a mistake on your original substitution $\endgroup$
    – podiki
    Commented Oct 30, 2023 at 13:50
  • $\begingroup$ I have also apied your way of approach@podiki. $\endgroup$
    – Sohini
    Commented Oct 30, 2023 at 13:52
  • $\begingroup$ If you are trying to tell any other approach then please tell me out@podiki $\endgroup$
    – Sohini
    Commented Oct 30, 2023 at 13:54

2 Answers 2

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$$\int \frac{1}{x(\ln x+1)}\,dx$$

Put $\ln x=t$ $\implies$ $\frac{1}{x}dx=dt$ $\implies$ $xdt=dx$

$$\int \frac{x}{x(t+1)}\,dt=\int \frac{1}{(t+1)}\,dt=\ln (t+1)+c=\ln(\ln x+1)+c$$

Where $c$ is the constant of integration.

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You have already received a good answer. But, I like to rectify your mistake.

"I am trying this question by substituting $\ln(x)=u$.So, $\frac{1}xdx=du$. Also $x$ will become $e^u$."

You are correct till this.

But,$$\int \frac{1}{x\ln x+x}dx\ne \color{#AA4A44}{\int \left[\frac{1}{(u+e^{u})}\right]du}$$

Because,

$$\require{cancel}\int \frac{1}{x\ln x+x}dx{=\int \frac{1}{e^u\ln e^u+e^u}(e^udu)\quad(\because \frac{1}xdx=du\Rightarrow dx=xdu=e^udu)\\=\int \frac{\color{#AA4A44}{\cancel{e^u}}}{\color{#AA4A44}{\cancel{e^u}}(\ln e^u+1)}du\\=\int \frac{1}{u+1}du\\=\ln|u+1|+c\\=\ln|\ln x+1|+c}$$

If you able to see the integrals,

$$\int \frac{1}{x\ln x+x}dx{=\int \frac{1}{(\ln x+1)}\cdot \frac{dx}{x}\\=\int \frac{d(\ln x +1)}{\ln x+1}}$$

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