=9∫ 1/(x+3)^(3/2)d(x+3)

=- 18/√(x+3)+c

So when x = 2,

- 18/√( 2+3)+c =- 18√5/5+c

X= infinity

- 18/√(∞+3)+c=c

So the original formula =-18 √ 5/5+c-c =-18 √ 5/5.