How do you find #int (x-1)/sqrt(x^2-2x)#?
1 Answer
Nov 10, 2016
Explanation:
#I=int(x-1)/sqrt(x^2-2x)dx=1/2int(2x-2)/sqrt(x^2-2x)dx#
Substituting
#I=1/2int(du)/sqrtu=1/2intu^(-1/2)du#
With the rule
#I=1/2u^(1/2)/(1/2)+C=u^(1/2)+C=sqrtu+C=sqrt(x^2-2x)+C#
