Question Number 194412 by SaRahAli last updated on 05/Jul/23
Answered by som(math1967) last updated on 05/Jul/23
$$\:\int{ln}\left({lnx}\right){dx}+\int\frac{{dx}}{\left({lnx}\right)^{\mathrm{2}} } \\ $$$$={ln}\left({lnx}\right)\int{dx}−\int\left\{\frac{{d}}{{dx}}{ln}\left({lnx}\right)\int{dx}\right\}{dx} \\ $$$$\:\:+\int\frac{{dx}}{\left({lnx}\right)^{\mathrm{2}} } \\ $$$$={xln}\left({lnx}\right)−\int\frac{{xdx}}{{xlnx}}\:+\int\frac{{dx}}{\left({lnx}\right)^{\mathrm{2}} } \\ $$$$={xln}\left({lnx}\right)−\frac{\mathrm{1}}{{lnx}}\int{dx}+\int\left\{\frac{{d}}{{dx}}×\frac{\mathrm{1}}{{lnx}}\right. \\ $$$$\left.\:\int{dx}\right\}{dx}+\int\frac{{dx}}{\left({lnx}\right)^{\mathrm{2}} } \\ $$$$={xln}\left({lnx}\right)−\frac{{x}}{{lnx}}\:−\int\frac{{dx}}{\left({lnx}\right)^{\mathrm{2}} }+\int\frac{{dx}}{\left({lnx}\right)^{\mathrm{2}} } \\ $$$$={xln}\left({lnx}\right)\:−\frac{{x}}{{lnx}}\:+{C} \\ $$