MHT CET202620 April 2026Evening ShiftMathematicsIndefinite IntegrationActual
If f(x)dx = g(x) + c then f⁻¹(x)dx =
Options
- Axf⁻¹(x) + c
- Bf(g⁻¹(x)) + c
- Cxf⁻¹(x) - g(f⁻¹(x)) + c
- Dg⁻¹(x) + c
Correct answer
C. xf⁻¹(x) - g(f⁻¹(x)) + c
Step-by-step solution
Let I = f⁻¹(x) dx Substitute f⁻¹(x) = t x = f(t) dx = f'(t) dt I = t f'(t) dt Applying integration by parts, taking t as the first function and f'(t) as the second function: I = t f'(t) dt - ( d dt (t) f'(t) dt ) dt I = t f(t) - f(t) dt Given f(x) dx = g(x) + c , we have 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