₀^ 4 ^2 x ^2 x ( ^3 x+ ^3 x )^2 ~d x=

Step-by-step solution

Let I = ₀^ 4 ^2 x ^2 x ( ^3 x + ^3 x)^2 d x . Dividing numerator and denominator by ^6 x gives: I = ₀^ 4 ^2 x ^2 x (1 + ^3 x)^2 d x Substituting u = 1 + ^3 x yields d u = 3 ^2 x ^2 x d x , so ^2 x ^2 x d x = 1 3 d u . The limits transform to u(0) = 1 and u( 4 ) = 2 . The integral becomes: I = 1 3 ₁^2 u⁻² d u = 1 3 [ - 1 u ]₁^2 = 1 3 ( - 1 2 + 1 ) = 1 3 1 2 = 1 6 Final answer: C