Question Number 110203 by Don08q last updated on 27/Aug/20
Answered by Dwaipayan Shikari last updated on 27/Aug/20
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)^{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+....=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}=\frac{{x}−\mathrm{1}}{{x}}=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$$$\int\mathrm{1}−\frac{\mathrm{1}}{{x}}{dx}={x}−{logx}+{C} \\ $$
Commented by Don08q last updated on 27/Aug/20
$${Thanks}\:{Sir} \\ $$