∫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

∫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

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$\int f (x) d x = g (x), t h e n \int f^{- 1} (x) d x =$

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$x f^{- 1} (x) - g {f^{- 1} (x)} + c$

$f^{- 1} (x) + c$

$g^{- 1} (x) + c$

$x f^{- 1} (x) - g^{- 1} (x) + c$

answer is B.

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Detailed Solution

$\begin{array}{l} \begin{array}{l} I = \int f^{- 1} (x) .1 d x \\ = f^{- 1} (x) \int d x - \int [\frac{d}{d x} f^{- 1} (x) \int d x] d x \\ = x f^{- 1} (x) - \int [x \frac{d}{d x} f^{- 1} (x)] d x \to (1) \\ L e t f^{- 1} (x) = t (\Rightarrow x = f (t)) \\ \Rightarrow d \{f^{- 1} (x)\} = d t \\ ∴ B y (1), w e h a v e \\ I = x f^{- 1} (x) - \int f (t) d t = x f^{- 1} (x) - g (t) [g i v e n \int f (x) d x = g (x)] \end{array} \\ = x f^{- 1} (x) - g \{f^{- 1} (x)\} + c \end{array}$