Ques. 1JEE Main 2024, 27 January (Evening Shift)
The integral (x8-x2 )dx (x12+3x6+1 ) -1 (x3+ 1x3 ) is equal to :
- e ( | -1 (x3+ 1x3 ) | )1/3+C
- e ( | -1 (x3+ 1x3 ) | )+C
- e ( | -1 (x3+ 1x3 ) | )1/2+C
- e ( | -1 (x3+ 1x3 ) | )3+C
Answer
(1) e ( | -1 (x3+ 1x3 ) | )1/3+C
Solution
Put t= -1 (x3+ 1x3 )
dt= 3 (x2- 1x4 )1+ (x3+ 1x3 )2 dx
= 3 (x6-1 )/x4 (x12+3x6+1 )/x6 dx
= 3 (x8-x2 )x12+3x6+1 dx
1+ (x3+ 1x3 )2= x12+3x6+1x6
I= 13 dtt= 13 |t|+C
= e | -1 (x3+ 1x3 ) |1/3+C
Answer: (1)