The value of the integral _ /6 ^ /3 ( 4 - ^2 x ^4 x ) dx is:

Step-by-step solution

The given integral is I = _ /6 ^ /3 ( 4 - ^2 x ^4 x ) dx . We can rewrite the integrand in terms of x and x . Using the identity ^2 x = 1 + ^2 x = 1 + 1 ^2 x , we get: I = _ /6 ^ /3 (4 - 1 - 1 ^2 x ) ^4 x , dx I = _ /6 ^ /3 (3 - 1 ^2 x ) (1 + ^2 x) ^2 x , dx Substitute t = x , which gives dt = ^2 x , dx . The limits of integration change as follows: When x = 6 , t = 1 3 . When x = 3 , t = 3 . The integral becomes: I = _ 1/ 3 ^ 3 (3 - 1 t^2 ) (1 + t^2) dt Expanding the integrand, we get: I = _ 1/ 3 ^ 3 (3 + 3t^2 - 1