Question
Integral sqrt(x-1/x^5)
Answers
Steve
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
∫√((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