Question

How do you find `int (x-1)/sqrt(x^2-2x)`?

Answer 1

`sqrt(x^2-2x)+C`

Explanation:

`I=int(x-1)/sqrt(x^2-2x)dx=1/2int(2x-2)/sqrt(x^2-2x)dx`

Substituting `u=x^2-2x` yields `du=(2x-2)dx`.

`I=1/2int(du)/sqrtu=1/2intu^(-1/2)du`

With the rule `intu^ndu=u^(n+1)/(n+1)+C`, we see that

`I=1/2u^(1/2)/(1/2)+C=u^(1/2)+C=sqrtu+C=sqrt(x^2-2x)+C`