Expert: Ahmed Salami - 9/18/2009

Question
intergrate, using the suggested substitution.
x over square root (x-3) and x=3+u^2

Answer
Hi Anna,
Using the substitution x = 3+u²
∫[x/√(x-3)]dx becomes ∫[(3+u²)/√(3+u²-3)]dx
= ∫[(3+u²)/√(u²)]dx
= ∫[(3+u²)/u]dx
= ∫[(3/u) + u]dx
but from x = 3+u², dx/du = 2u and u = √(x-3)
dx = 2u du
The integral becomes ∫[(3/u) + u]. 2u du
= ∫[6 + 2u²]du
= 6u + 2u³/3 + c
= (2u/3)(9 + u²) + c
= [2√(x-3)/3].[9 + (x-3)] + c
= [2√(x-3)/3].(x+6) + c

Regards