JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given integral is I = _ /6 ^ /4 ( (x- 3 ) (x+ 3 )+1 )dx Using the identity A B + 1 = (A-B) A B , the integrand simplifies to: f(x) = ( (x- 3 ) - (x+ 3 ) ) (x- 3 ) (x+ 3 ) f(x) = (- 2 3 ) ^2 x - ^2 ( 3 ) f(x) = -1/2 ^2 x - 3/4 = 2 3 - 4 ^2 x Dividing the numerator and the denominator by ^2 x: f(x) = 2 ^2 x 3 ^2 x - 4 ^2 x = 2 ^2 x 3(1+ ^2 x) - 4 ^2 x = 2 ^2 x 3 - ^2 x The integral becomes: I = _ /6 ^ /4 2 ^2 x 3 - ^2 x dx Substituting x = t ^2 x dx = dt. When x = 6 , t = 1 3 . When x = 4 , t = 1. I = _ 1/ 3 ^ 1 2 3 - t^2 dt Using the standard integral 1 a^2 - x^2 dx = 1 2a | a+x a-x |: I = 2 [ 1 2 3 | 3 +t 3 -t | ]_ 1/ 3 ^ 1 I = 1 3 ( ( 3 +1 3 -1 ) - ( 3 + 1 3 3 - 1 3 ) ) I = 1 3 ( ( 3 +1 3 -1 ) - ( 3+1 3-1 ) ) I = 1 3 ( ( 3 +1 3 -1 ) - 2 ) = 1 3 ( 3 +1 2( 3 -1) ) Rationalizing the term inside the logarithm by multiplying the numerator and the denominator by ( 3 -1): 3 +1 2( 3 -1) = ( 3 +1)( 3 -1) 2( 3 -1)^2 = 3-1 2( 3 -1)^2 = 1 ( 3 -1)^2 = ( 3 -1)^ -2 Thus, I = 1 3 (( 3 -1)^ -2 ) = - 2 3 ( 3 -1) Comparing this with I = _e( 3 -1), we get: = - 2 3 Therefore, 9 ^2 = 9 (- 2 3 )^2 = 9 4 3 = 12. Answer: 12