If |
|||
(a) |
−4x3−1 |
(b) |
4x3+1 |
(c) |
−2x3−1 |
(d) |
−2x3+1 |
If |
|||
(a) |
−4x3−1 |
(b) |
4x3+1 |
(c) |
−2x3−1 |
(d) |
−2x3+1 |
f(x)+c
Put x3=t
3x2 dx=dt
dt
[4t+1]+c
[4x3+1]+c
∴f(x)= −1−4x3
Option(a)
(From the given option (a)is most suitable)