Please help me integrate the following....?

Question

the integral is.....

3dx / x^2+6x+10

the 3dx is over the rest of the equation, i know you have to complete the square on the bottom first, which will be (x+3)^2 + 1. I need to use the substitution method where you let something equal u, then find du. Please help! I have a test in about 8 hours and need to know this.

Answer 1

3dx/(x+3)^2 + 1

let x+3 = u
dx=du

3du/(1+u^2)

answer: 3 arctanu +C
or
3arctan(x+3)+C

Answer 2

When you have (something)^2 + 1 in a nasty location, such as a denominator or inside a square root, often the best subsitution to make is (something) = tan θ, due to the identity tan^2 θ + 1 = sec^2 θ. (If it's a constant other than 1, scale appropriately: e.g. for x^2 + 4 substitute x = 2 tan θ to get 4tan^2 θ + 4 = 4 sec^2 θ.)

So here we let x+3 = tan θ, so dx = sec^2 θ dθ and we get
∫(3 sec^2 θ dθ) / (tan^2 θ + 1)
= ∫(3 sec^2 θ dθ) / (sec^2 θ)
= ∫3 dθ
= 3θ + c
= 3 arctan (x+3) + c.

Answer 3

∫1/(1+u²) du = arctan(u)

∫3dx/(x² + 6x + 10)
= ∫3/[(x + 3)² + 1] dx
Let u = (x+3)
du = dx

∫3/[(x + 3)² + 1] dx
= ∫3/(1+u²) du
= 3arctan(u) + c
= 3arctan(x+3) + c