
William S. answered 01/29/14
Tutor
4.4
(10)
Experienced scientist, mathematician and instructor - William
∫ x(x2 + 2)2dx
Let u = x2 + 2
du = 2xdx
∫ x(x2 + 2)2dx = (1/2)∫u2du = (1/2)*(u3/3) = (1/6)*(x2 + 2)3 (evaluated between 0 and 1)
= (1/6)[(1)2 + 2]3 - (1/6)[(0)2 + 2]3 = (1/6)*(27 - 8) = (19/6)
Sun K.
01/29/14