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

Correct Answer is : xf−1(x)−g(f−1(x))+cx f^{-1}(x)-g≤ft(f^{-1}(x))+cxf−1(x)−g(f−1(x))+c

x f^{-1}(x)-g≤ft(f^{-1}(x))+c

Correct Answer is : xf−1(x)−g(f−1(x))+cx f^{-1}(x)-g≤ft(f^{-1}(x))+cxf−1(x)−g(f−1(x))+c

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MHTCET 20 Apr 2026 Shift 2 Paper

Question : 128 of 150

Marks: +1, -0

\int f(x) d x=g(x)+c

\int f^{-1}(x) d x=

Solution:

Concept:

The integral of an inverse function can be found by substitution and integration by parts.

Explanation:

Given

\int f(x)\,dx = g(x) + c

, we know

g'(x) = f(x)

Let

y = f^{-1}(x)

. Then

x = f(y)

, so

dx = f'(y)\,dy

Thus,

\int f^{-1}(x)\,dx = \int y\,f'(y)\,dy

Apply integration by parts with

u=y

and

dv = f'(y)\,dy

We get

y\,f(y) - \int f(y)\,dy

Since

y = f^{-1}(x)

and

f(y) = x

, we have

y\,f(y) = x f^{-1}(x)

Also,

\int f(y)\,dy = g(y) + c = g(f^{-1}(x)) + c

Therefore,

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

Answer:

Option C:

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

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