Question

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

• A. \(xf^{-1}(x)+c\)
• B. \(f(g^{-1}(x))+c\)
• C. \(xf^{-1}(x)-g(f^{-1}(x))+c\)
• D. \(g^{-1}(x)+c\)

Hint:

Substitute x = f(y) so that the inverse function becomes y, then integrate by parts.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Let \(y=f^{-1}(x)\), so \(x=f(y)\) and \(dx=f'(y)\,dy\).

Step 2: Substitute
\[ \int f^{-1}(x)\,dx=\int y\,f'(y)\,dy \]

Step 3: Integrate by parts
Take \(u=y\) and \(dv=f'(y)\,dy\):
\[ \int y f'(y)\,dy=y f(y)-\int f(y)\,dy \]

Step 4: Use the given information
\(\int f(y)\,dy=g(y)+c\). Also \(f(y)=x\) and \(y=f^{-1}(x)\). So
\[ \int f^{-1}(x)\,dx=x f^{-1}(x)-g\big(f^{-1}(x)\big)+c \]
This is option (C). Option (A) forgets the subtraction of \(g\).

Final Answer:
The integral of the inverse function is \(xf^{-1}(x)-g(f^{-1}(x))+c\), option (C). \[ \boxed{xf^{-1}(x)-g(f^{-1}(x))+c} \]