MHT CET202617 April 2026Evening ShiftMathematicsIndefinite IntegrationActual
If f(x) = ⁻¹x 1 - x^2 and g(x) = e^ ⁻¹x , then the value of f(x)g(x) , dx =
Options
- Ae^ ⁻¹x ( ⁻¹x - 1) + c
- Be^ ⁻¹x (1 - ⁻¹x) + c
- Ce^ ⁻¹x ( ⁻¹x + 1) + c
- De^ ⁻¹x (- ⁻¹x - 1) + c
Correct answer
A. e^ ⁻¹x ( ⁻¹x - 1) + c
Step-by-step solution
Let I = f(x)g(x) , dx = ⁻¹x 1 - x^2 e^ ⁻¹x , dx Substitute t = ⁻¹x , which gives dt = 1 1 - x^2 , dx The integral becomes: I = t e^t , dt Using integration by parts, we get: I = t e^t - e^t , dt I = t e^t - e^t + c I = e^t(t - 1) + c Substituting t = ⁻¹x back into the expression: I = e^ ⁻¹x ( ⁻¹x - 1) + c Answer: e^ ⁻¹x ( ⁻¹x - 1) + c