Q91 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
If $\int f(x)dx = g(x) + c$ then $\int f^{-1}(x)dx =$
- A$xf^{-1}(x) + c$
- B$f(g^{-1}(x)) + c$
- Correct answerC$xf^{-1}(x) - g(f^{-1}(x)) + c$
- D$g^{-1}(x) + c$
Solution
Let $I = \int f^{-1}(x) dx$
Substitute $f^{-1}(x) = t \Rightarrow x = f(t) \Rightarrow dx = f'(t) dt$
$I = \int t f'(t) dt$
Applying integration by parts, taking $t$ as the first function and $f'(t)$ as the second function:
$I = t \int f'(t) dt - \int \left( \dfrac{d}{dt}(t) \int f'(t) dt \right) dt$
$I = t f(t) - \int f(t) dt$
Given $\int f(x) dx = g(x) + c$, we have $\int f(t) dt = g(t) + c$
$I = t f(t) - g(t) + C$
Substituting $t = f^{-1}(x)$ and $f(t) = x$:
$I = x f^{-1}(x) - g(f^{-1}(x)) + C$
Answer: $xf^{-1}(x) - g(f^{-1}(x)) + c$
Substitute $f^{-1}(x) = t \Rightarrow x = f(t) \Rightarrow dx = f'(t) dt$
$I = \int t f'(t) dt$
Applying integration by parts, taking $t$ as the first function and $f'(t)$ as the second function:
$I = t \int f'(t) dt - \int \left( \dfrac{d}{dt}(t) \int f'(t) dt \right) dt$
$I = t f(t) - \int f(t) dt$
Given $\int f(x) dx = g(x) + c$, we have $\int f(t) dt = g(t) + c$
$I = t f(t) - g(t) + C$
Substituting $t = f^{-1}(x)$ and $f(t) = x$:
$I = x f^{-1}(x) - g(f^{-1}(x)) + C$
Answer: $xf^{-1}(x) - g(f^{-1}(x)) + c$
Hence, the correct answer is option: 'C'.