If ∫f(x)dx=g(x)+c, then ∫f−1(x)dx=

put x= f(t) dx=f'(t)dt∫f-1(x) dx=∫t.f-1(t) dt=t ft-∫ 1 ftdt by parts=f-1xx-gt+c =xf-1x-gf-1x+c

If ∫f(x)dx=g(x)+c, then ∫f−1(x)dx=

put x= f(t) dx=f'(t)dt∫f-1(x) dx=∫t.f-1(t) dt=t ft-∫ 1 ftdt by parts=f-1xx-gt+c =xf-1x-gf-1x+c

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$If \int f (x) d x = g (x) + c, then \int f^{- 1} (x) d x =$

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

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

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

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

answer is C.

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Detailed Solution

put x= f(t)
dx=f'(t)dt

$\int f^{- 1} (x) d x = \int t . f^{- 1} (t) d t$
$= t f (t) - \int_{} (1) f (t) d t (b y p a r t s)$
$= f^{- 1} (x) x - g (t) + c = x f^{- 1} (x) - g (f^{- 1} (x)) + c$

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If ∫f(x)dx=g(x)+c, then ∫f−1(x)dx=

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$If \int f (x) d x = g (x) + c, then \int f^{- 1} (x) d x =$