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

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