If `intf(x)dx=g(x)` then `intf^(-1)(x) dx` is

Options

A. `xf^(-1)(x)+C`
B. `f{g^(-1)(x)}+C`
C. `xf^(-1)(x)-g{f^(-1)(x)}+C`
D. `g^(-1)(x)+C`

Answer

C

Explanation

Let `f^(-1)(x)=u rArr x=f(u)`
`rArr" "dx=f'(u)du`
`therefore" "l=int f^(-1)(x)dx=intuf'(u)du`
`rArr" "l=u f(u)-intf(u)du=uf(u)-g(u)+C`
`" "[because int f(x)dx=g(x)+C," given"]`
`therefore` On putting `u=f^(-1)(x) and f(u)=x,` we get
`l=xf^(-1)(x)-g{f^(-1)(x)}+C`