MHT CET202619 April 2026Evening ShiftMathematicsIndefinite IntegrationActual
The value of integral x^3 x ,dx is...
Options
- Ax^3 x + x^2 x - 6x x + 6 x + c
- Bx^3 x + 3x^2 x - 6x x - 6 x + c
- Cx^3 x + 3x^2 x - 6x x - 6 x + c
- Dx^3 x + 3x^2 x - 6x x + 6 x + c
Correct answer
C. x^3 x + 3x^2 x - 6x x - 6 x + c
Step-by-step solution
Let I = x^3 x , dx . Using integration by parts, u , dv = uv - v , du . Taking u = x^3 and dv = x , dx , we get: I = x^3 x - 3x^2 x , dx Applying integration by parts again for 3x^2 x , dx with u = 3x^2 and dv = x , dx : I = x^3 x - [ 3x^2 (- x) - 6x (- x) , dx ] I = x^3 x + 3x^2 x - 6x x , dx Applying integration by parts once more for 6x x , dx with u = 6x and dv = x , dx : I = x^3 x + 3x^2 x - [ 6x x - 6 x , dx ] I = x^3 x + 3x^2 x - 6x x + 6 x , dx I = x^3 x + 3x^2 x - 6x x - 6 x + c Answer: x^3 x + 3x^2 x - 6x