Calculus
posted by Anon .
Evaluate integral of x times the square root of (x4) dx
I got u(8u^(3/2) over 3)
Can anyone check this for me?

if u^2 = x4
2u du = dx
x = u^2+4
∫x√(x4) dx = ∫(u^2+4)u 2u du
= ∫2u^4 + 8u^2 du
= 2/5 u^5 + 8/3 u^3
= u^3 (2/5 u^2 + 8/3)
= (x4)^(3/2) (2/5 (x4) + 8/3)
= 2/15 (x4)^(3/2) (3x+8)