The value of the integral 02 x (x2+x+1 ) ( x+1 ) ( x4+x2+1 ) dx is equal to :
- 13 e (3-2 2 )
- 23 e (4+ 2 )
- 23 e (3+2 2 )
- 13 e (1+6 2 )
Answer
(3) 23 e (3+2 2 )
Solution
x4+x2+1= (x2+x+1 ) (x2-x+1 ) and (x+1) (x2-x+1 )=x3+1
integrand = x x2+x+1 x+1 x2+x+1 x2-x+1= x 1+x3
Put x3/2=u 32 x dx=du; x: 0 2 u: 0 2 2
I= 23 02 2 du 1+u2= 23 [ e (u+ 1+u2 ) ]02 2
= 23 [ e (2 2+3 )- e1 ]= 23 e (3+2 2 )
Answer: (3)