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Question

If  f is an invertible function such that f(x) dx=g(x) for some function g, then f1(x) dx is equal to


Correct Answer
A
xf1(x)g(f1(x))+C, where C is a constant.
 
Your Answer
B
f1(x)xg(f1(x))+C, where C is a constant.
Your Answer
C
(x1)f1(x)+xg(f1(x))+C, where C is a constant.
Your Answer
D
x22f1(x)+g(f1(x))+C, where C is a constant.

Solution

The correct options are
A xf1(x)g(f1(x))+C, where C is a constant.
 
We know that,
f(x)=g(x)
I=f1(x) dx
Assuming f1(x)=t
f(t)=xf(t)dt=dxg′′(t)dt=dx

I=f1(x) dxI=tg′′(t)dtI=tg(t)g(t)dtI=tg(t)g(t)+CI=f1(x)f(t)g(f1(x))+CI=xf1(x)g(f1(x))+C

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