Question Number 216381 by glory86 last updated on 06/Feb/25
$$\int\left({lnx}\right)^{\mathrm{2}} {dx} \\ $$
Answered by mr W last updated on 06/Feb/25
![∫(ln x)^2 dx =x(ln x)^2 −∫x(2 ln x)(1/x)dx =x(ln x)^2 −2∫ (ln x)dx =x(ln x)^2 −2[x(ln x)−∫x×(1/x)dx] =x(ln x)^2 −2x(ln x)+2x+C =x(ln x−1)^2 +x+C](https://www.tinkutara.com/question/Q216384.png)
$$\int\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} {dx} \\ $$$$={x}\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} −\int{x}\left(\mathrm{2}\:\mathrm{ln}\:{x}\right)\frac{\mathrm{1}}{{x}}{dx} \\ $$$$={x}\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} −\mathrm{2}\int\:\left(\mathrm{ln}\:{x}\right){dx} \\ $$$$={x}\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} −\mathrm{2}\left[{x}\left(\mathrm{ln}\:{x}\right)−\int{x}×\frac{\mathrm{1}}{{x}}{dx}\right] \\ $$$$={x}\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} −\mathrm{2}{x}\left(\mathrm{ln}\:{x}\right)+\mathrm{2}{x}+{C} \\ $$$$={x}\left(\mathrm{ln}\:{x}−\mathrm{1}\right)^{\mathrm{2}} +{x}+{C} \\ $$