₁^3 x^2 (16 x^2-8 x^3+x^4 ) d x=

Step-by-step solution

Given integral: I = ₁^3 x^2 (16 x^2 - 8 x^3 + x^4) , dx Simplify the integrand. The numerator is x^2 = 2 x . The denominator’s argument factors as 16x^2 - 8x^3 + x^4 = x^2(16 - 8x + x^2) = x^2(x - 4)^2 , so (x^2(x - 4)^2) = 2 x + 2 |x - 4| . Since x [1, 3] , x - 4 Substituting, I = ₁^3 2 x 2 x + 2 (4 - x) , dx = ₁^3 x x + (4 - x) , dx . Apply the substitution x 4 - x to the integral, giving: I = ₁^3 (4 - x) (4 - x) + x , dx . Adding these two expressions for I : 2I = ₁^3 ( x x + (4 - x) + (4 - x) (4 - x) + x ) dx =