∫f(x) dx = x^2+C


∫(1/x^2) f(1/x) dx
let
y= 1/x
dy = -(1/x^2)dx
∫(1/x^2) f(1/x) dx
= -∫ f(y)dy
= -y^2 + C
= -1/x^2 + C