Answer
$\frac{1}{2}\ln|\frac{x-2}{x}|+c$
Work Step by Step
Ɪ=$\int\frac{dx}{x^{2}-2x}$
by completing the square
Ɪ=$\int\frac{dx}{(x^{2}-2x+1)-1}$
Ɪ=$\int\frac{dx}{(x-1)^{2}-1}$
using formula 6 :
$\int\frac{dx}{x^{2}-a^{2}}$=$\frac{1}{2a}\ln|\frac{x-a}{x+a}|+c$
then the integral becomes
Ɪ=$\int\frac{dx}{((x-1)^{2})-1}$=$\frac{1}{2}\ln|\frac{(x-1)-1}{(x-1)+1}|+c$
Ɪ=$\frac{1}{2}\ln|\frac{x-2}{x}|+c$
