Evaluate the definite integral?

Question

Int[x * sqrt(x+2)]
I end up with 1/du after substituting and it is not working.
Help!!

Answer 1

Integration by parts

u = x
dv = sqrt(x +2) dx

du = dx
v = (2/3)(x+2)^(3/2)

uv - int(v du)
= x (2/3)(x+2)^(3/2) - int((2/3)(x+2)^(3/2) dx)
= (2x/3)(x+2)^(3/2) - (2/3)int((x+2)^(3/2) dx)
= (2x/3)(x+2)^(3/2) - (2/3)[(2/5)(x+2)^(5/2)] + C
= (2x/3)(x+2)^(3/2) - (4/15)(x+2)^(5/2) + C

Answer 2

Alternatively, you could use the substitution

u = x+2, so x = u-2, dx = du and get

integral[(u-2)sqrt(u)du]
= integral [(u^(3/2) - 2u^(1/2) )du]

etc.

Answer 3

Integration by Parts

x = u

sqrt (x+2) dx = dv

dx = du

(2/3)(x+2)^3/2 = v

(2/3)(x)(x+2)^3/2 - Integral (2/3)(x+2)^3/2 dx

2/3 ((x)(x+2)^3/2 - Integral (x+2)^3/2 dx)

2/3 ((x)(x+2)^3/2 - (2/5)(x+2)^5/2) + C)
_________________________

Why do you have to answer first T-T....