∫f(x)dx=g(x), then ∫ f-1(x) dx =
g-1(x)+c
xf-1(x)-g{f-1(x)} +c
xf-1(x)-g-1(x) +c
f-1(x)+c
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