Evaluate this definite integral

2x ∜x^4+8x^2+16 limits 3,√5

2x ∜x^4+8x^2+16 dx
= sqrt(x^2+4) (x^2+4)' dx
= (2/3)(x^2+4)^(3/2)
= (2/3)[(3^2+4)^(3/2) - (5+4)^(3/2)]
= (2/3)[13^1.5 - 27]
= 13.248

∫2x (x^4 + 8x^2 + 16)^(1/4) dx from [√5 to 3]

= ∫2x (x^2 + 4)(1/2) dx

let x^2 + 4 = u^2 , when x = 3, u = √13 and when x = √5, u = 3
2x dx = 2u du

∫ 2 u (u) du from [ 3 to √13 ]

= u^2

= [13 - 9] = 4