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Integrals · Question 1

JEE Main 2026 (Online) 8th April Evening ShiftMCQMedium📈 Topic in 13% papers+4−1
1

The value of the integral 02 x (x2+x+1 ) ( x+1 ) ( x4+x2+1 ) dx is equal to :

  1. 13 e (3-2 2 )
  2. 23 e (4+ 2 )
  3. 23 e (3+2 2 )
  4. 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)

JEE Main 2026 (Online) 8th April Evening Shift

Keep practising — Integrals

2

If /6 /4 ( (x- 3 ) (x+ 3 )+1 )dx = e ( 3-1 ), then 9 2 is equal to .

JEE Main 2026 (Online) 8th April Evening Shift
3

The value of the integral - /4 /4 ( 32 4 x1+e x )dx is :

  1. 4 +2
  2. 3 +8
  3. 3 +4
  4. 4 +3
JEE Main 2026 (Online) 6th April Morning Shift
4

The value of the integral -11 ( x3+|x|+1x2+2|x|+1 )dx is equal to :

  1. 3 e 2
  2. 2 e 2
  3. 5 e 3
  4. 3 e 3
JEE Main 2026 (Online) 6th April Evening Shift
5

The value of the integral 0 e(x)x2+4 dx is:

  1. e(2)2
  2. e(2)4
  3. 1+ e(2)
  4. 2+ e(2)
JEE Main 2026 (Online) 5th April Morning Shift
6

The value of the integral 6 3 ( 4-cosec2 x 4 x )dx is :

  1. 11 3
  2. 16 3
  3. 323 3
  4. 643 3
JEE Main 2026 (Online) 5th April Morning Shift
7

Let (21-a+21+a ), f(a), (3a+3-a ) be in A.P. and be the minimum value of f(a). Then the value of the integral e( -1) e( ) dx (e2x-e-2x ) is :

  1. 12 e ( 43 )
  2. 14 e ( 43 )
  3. 12 e ( 85 )
  4. 14 e ( 85 )
JEE Main 2026 (Online) 5th April Evening Shift
8

Let f : [1, ) R be a differentiable function defined as f(x)= 1x f(t) dt +(1-x) ( e x-1 )+e. Then the value of f(f(1)) is :

  1. (1+ee )
  2. (1+e)
  3. (1+e+ee )
  4. 1+2e
JEE Main 2026 (Online) 5th April Evening Shift
9

Let -22 (| x|+[x x] )dx=2(3- 2)+ , where [ ] is the greatest integer function. Then ( 2 ) equals:

  1. 1
  2. 2
  3. 4
  4. 8
JEE Main 2026 (Online) 4th April Morning Shift
10

If =1 and =1+i 2, where i= -1 are two roots of the equation x3+ax2+bx+c=0, a, b, c R, then -11 (x3+ax2+bx+c )dx is equal to :

  1. -2
  2. -4
  3. -8
  4. -10
JEE Main 2026 (Online) 4th April Evening Shift
11

The integral 01 -1 (1+x+x2 )dx is equal to :

  1. 2 -12+ 12 e ( 54 )+ 2
  2. 2 -12+ 12 e ( 54 )- 2
  3. 2 -12- 12 e ( 54 )+ 2
  4. 2 -12- 12 e ( 54 )- 2
JEE Main 2026 (Online) 4th April Evening Shift

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