Question

# If  f is an invertible function such that ∫f(x) dx=g(x) for some function g, then ∫f−1(x) dx is equal to

A
xf1(x)g(f1(x))+C, where C is a constant.

B
f1(x)xg(f1(x))+C, where C is a constant.
C
(x1)f1(x)+xg(f1(x))+C, where C is a constant.
D
x22f1(x)+g(f1(x))+C, where C is a constant.

Solution

## The correct options are A xf−1(x)−g(f−1(x))+C, where C is a constant.  We know that, f(x)=g′(x) I=∫f−1(x) dx Assuming f−1(x)=t f(t)=x⇒f′(t)dt=dx⇒g′′(t)dt=dx I=∫f−1(x) dx⇒I=∫tg′′(t)dt⇒I=tg′(t)−∫g′(t)dt⇒I=tg′(t)−g(t)+C⇒I=f−1(x)f(t)−g(f−1(x))+C⇒I=xf−1(x)−g(f−1(x))+C

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