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Question Number 110203 by Don08q last updated on 27/Aug/20

Answered by Dwaipayan Shikari last updated on 27/Aug/20

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{1}{1-x}\right)}^n=1+\frac{1}{1-x}+\frac{1}{{(1-x)}^2}+....=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}=1-\frac{1}{x}$$$$\int 1-\frac{1}{x}dx=x-logx+C$$

Commented by Don08q last updated on 27/Aug/20

$$ThanksSir$$

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