If _ /6 ^ 5 /6 x (4- ^2 x)(1+2^ x ) dx = 1 ( + 3 ) , where , , are positive integers, then the value of + + is equal to

Step-by-step solution

Let I = _ /6 ^ 5 /6 x (4- ^2 x)(1+2^ x ) dx Using the property _a^b f(x) dx = _a^b f(a+b-x) dx , we replace x with - x . Since ( - x) = x and ( - x) = - x , we get: I = _ /6 ^ 5 /6 x (4- ^2 x)(1+2^ - x ) dx I = _ /6 ^ 5 /6 2^ x x (4- ^2 x)(1+2^ x ) dx Adding the two expressions for I gives: 2I = _ /6 ^ 5 /6 x (1 + 2^ x ) (4- ^2 x)(1+2^ x ) dx = _ /6 ^ 5 /6 x 4- ^2 x dx Let x = t , then - x dx = dt . When x = /6 , t = 3 /2 . When x = 5 /6 , t = - 3 /2 . 2I = _ - 3 /2 ^ 3 /2 dt 4-t^2 = 2 ₀^ 3 /2 dt 4-t^2 I = ₀^ 3 /2