Question Number 170802 by Sotoberry last updated on 31/May/22
Answered by thfchristopher last updated on 31/May/22
![∫ln (x+x^2 )dx =∫ln x(x+1)dx =∫ln xdx+∫ln (x+1)dx =xln x−∫xd(ln x)+xln (x+1)−∫xd[ln (x+1)] =xln x−∫dx+xln (x+1)−∫(x/(x+1))dx =xln x−x+xln (x+1)−∫(1−(1/(x+1)))dx =xln x−x+xln (x+1)−∫dx+∫(1/(x+1))dx =xln x−2x+(x+1)ln (x+1)+C =ln x^x (x+1)^(x+1) −2x+C](https://www.tinkutara.com/question/Q170804.png)
$$\int\mathrm{ln}\:\left({x}+{x}^{\mathrm{2}} \right){dx} \\ $$$$=\int\mathrm{ln}\:{x}\left({x}+\mathrm{1}\right){dx} \\ $$$$=\int\mathrm{ln}\:{xdx}+\int\mathrm{ln}\:\left({x}+\mathrm{1}\right){dx} \\ $$$$={x}\mathrm{ln}\:{x}−\int{xd}\left(\mathrm{ln}\:{x}\right)+{x}\mathrm{ln}\:\left({x}+\mathrm{1}\right)−\int{xd}\left[\mathrm{ln}\:\left({x}+\mathrm{1}\right)\right] \\ $$$$={x}\mathrm{ln}\:{x}−\int{dx}+{x}\mathrm{ln}\:\left({x}+\mathrm{1}\right)−\int\frac{{x}}{{x}+\mathrm{1}}{dx} \\ $$$$={x}\mathrm{ln}\:{x}−{x}+{x}\mathrm{ln}\:\left({x}+\mathrm{1}\right)−\int\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){dx} \\ $$$$={x}\mathrm{ln}\:{x}−{x}+{x}\mathrm{ln}\:\left({x}+\mathrm{1}\right)−\int{dx}+\int\frac{\mathrm{1}}{{x}+\mathrm{1}}{dx} \\ $$$$={x}\mathrm{ln}\:{x}−\mathrm{2}{x}+\left({x}+\mathrm{1}\right)\mathrm{ln}\:\left({x}+\mathrm{1}\right)+{C} \\ $$$$=\mathrm{ln}\:{x}^{{x}} \left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} −\mathrm{2}{x}+{C} \\ $$
Answered by Mathspace last updated on 31/May/22
$${by}\:{parts}\:{I}={xln}\left({x}+{x}^{\mathrm{2}} \right)−\int{x}×\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+{x}^{\mathrm{2}} }{dx} \\ $$$$={xln}\left({x}+{x}^{\mathrm{2}} \right)−\int\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+\mathrm{1}}{dx} \\ $$$$={xln}\left({x}+{x}^{\mathrm{2}} \right)−\int\:\frac{\mathrm{2}\left({x}+\mathrm{1}\right)−\mathrm{1}}{{x}+\mathrm{1}}{dx} \\ $$$$={xln}\left({x}+{x}^{\mathrm{2}} \right)−\mathrm{2}{x}+{ln}\mid{x}+\mathrm{1}\mid\:+{c} \\ $$