Integral sqrt(x-1/x^5)

Answer 1

You have

∫√((x-1)/x^5) dx
= ∫√((x-1)/x) 1/x^2 dx

If you let
u = (x-1)/x
du = 1/x^2 dx

and you have

∫√u du
= (2/3) u^(3/2)
= (2/3) ((x-1)/x)^(3/2) + C