∫f(x)dx=g(x), then ∫ f-1(x) dx =

I=∫ f−1x.1 dx​    = f−1x ∫dx−∫ddx f−1x ∫dx dx​   =xf−1x −∫xddx f−1x dx​        →1Let   f−1x = t   ⇒x=ft⇒ ​df−1x=dt​∴  By 1,we haveI=xf−1x−∫ftdt​=xf−1x−gt   given ∫fxdx=gx=xf−1x−gf−1x +c