JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

I = _ 0 ^ 1 ^ -1 (1+x+x^2)dx I = _ 0 ^ 1 ^ -1 ( 1 1+x+x^2 )dx I = _ 0 ^ 1 ^ -1 ( (x+1)-x 1+(x+1)x )dx I = _ 0 ^ 1 ( ^ -1 (x+1)- ^ -1 x)dx I = _ 0 ^ 1 ^ -1 (x+1)dx - _ 0 ^ 1 ^ -1 x dx Substituting x+1 = t in the first integral: I = _ 1 ^ 2 ^ -1 x dx - _ 0 ^ 1 ^ -1 x dx Using integration by parts, ^ -1 x dx = x ^ -1 x - 1 2 _e(1+x^2) I = [x ^ -1 x - 1 2 _e(1+x^2) ]_ 1 ^ 2 - [x ^ -1 x - 1 2 _e(1+x^2) ]_ 0 ^ 1 I = (2 ^ -1 2 - 1 2 _e 5 - ( 4 - 1 2 _e 2 ) ) - ( 4 - 1 2 _e 2 - 0 ) I = 2 ^ -1 2 - 1 2 _e 5 - 2 + _e 2 I = 2 ^ -1 2 - 1 2 _e 5 + 1 2 _e 4 - 2 I = 2 ^ -1 2 - 1 2 _e ( 5 4 ) - 2 Answer: 2 ^ -1 2- 1 2 _e ( 5 4 )- 2