∫$$f(x)dx=g(x)$$, then ∫ $$f-1(x)$$ dx $$=$$

Correct option is 2xf-1(x)-g{f-1(x)} +c

Questions

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

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g-1(x)+c

xf-1(x)-g{f-1(x)} +c

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detailed solution

Correct option is B

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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∫f(x)dx=g(x), then ∫ f-1(x) dx =

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free study material

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