# How to integrate $\int \frac{1}{x\ln x+x}dx$?

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.

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

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

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.

"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,

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