Integrate x / (sqrt(x+3)) dx from 1 to 6

http://www.wolframalpha.com/input/?i=int…

that's the answer but I want to know the steps on how to do it, I heard substitution works but I don't know how. 10 points fast :)

∫ x/√(x+3) dx

let u = x + 3
x = u - 3
du = dx

∫ (u - 3)/√u du
∫ (u - 3)*u^(-1/2) du
∫ (u^(1/2) - 3u^(-1/2)) du
(2/3)u^(3/2) - 6u^(1/2)
(2/3)(x+3)^(3/2) - 6(x+3)^(1/2) | (1,6)

Evaluate at 6 and 1, then subtract the two to get your answer.

Possible intermediate steps:
integral x/sqrt(3+x) dx
For the integrand x/sqrt(x+3), substitute u = x+3 and du = dx:
= integral (u-3)/sqrt(u) du
Expanding the integrand (u-3)/sqrt(u) gives sqrt(u)-3/sqrt(u):
= integral (sqrt(u)-3/sqrt(u)) du
Integrate the sum term by term and factor out constants:
= integral sqrt(u) du-3 integral 1/sqrt(u) du
The integral of 1/sqrt(u) is 2 sqrt(u):
= integral sqrt(u) du-6 sqrt(u)
The integral of sqrt(u) is (2 u^(3/2))/3:
= (2 u^(3/2))/3-6 sqrt(u)+constant
Substitute back for u = x+3:
= 2/3 (x+3)^(3/2)-6 sqrt(x+3)+constant
Which is equal to:
= 2/3 (x-6) sqrt(x+3)+constant

x / (sqrt(x+3)) dx

let , x+3 = v

now dv/dx = d/dx(x+3) = 1

=> dv=dx

again , x = v-3

now, ∫ x / (sqrt(x+3)) dx

= ∫ (v-3)/sqrt v dv

= ∫ (v-3)*(v^(-1/2)) dv

= ∫ [ v*v^(-1/2) - 3*v^(-1/2) ] dv

= ∫ [ v^(1/2) - 3*v^(-1/2) ] dv

= (2/3)*v^(3/2) - 6v^(1/2) +c

= (2/3)(x+3)^(3/2) - 6(x+3)^(1/2) +c

from (1) to (6)

= 12 +c Ans.