Can anyone help with Algebraic Manipulation?

Question

How do I algebraically manipulate

(u+2) / (u^2+u+1) to become 1 / [((u+1/2)^2) + 3/4] ?

I need to get (u+2) / (u^2+u+1) into the form 1 / (a^2 + b^2) so I can use arctan to integrate.

One suggestion I was given was that :
(u+2)/(u^2+u+1) = 1/2(2u+1)/(u^2+u+1)+3/2(u^2+u+...

I completed the square for the denominator and was able to get the sum of 2 quantities squared. I can't figure out how to get the denominator to just 1.

Answer 1

The suggestion you were given was good. Once you break this apart, you can integrate 1/2(2u+1)/(u^2+u+1) using the substitution v=u²+u+1, giving you ln |u²+u+1| (the 2u+1 cancels after the substitution).

For the 3/2/(u²+u+1), consider the following manipulation:
3/2 / (u²+u+1/4+3/4) (break apart the 1)
3/2 / ((u+1/2)²+3/4) (factor the left)
Then use the substitution √3/2 tan v=u+1/2 to get:
3/2/(3/4(tan² v + 1)) sec² v dv
2 dv
Which when integrated becomes:
2v
2 arctan ((u+1/2)/(√3/2))
2 arctan ((2u+1)/√3)

Answer 2

You don't.

You write the original expression as
(u+1/2 +3/2) / (u^2 + u + 1) =
(u+1/2) / (u^2 + u + 1) + (3/2) / (u^2 + u+ 1).

For the second term, complete the square like you did earlier, and you can use arctan to integrate.

For the first term, write it as (2u+1) / [2(u^2+u+1)]. Then use a substitution so you can use ln to integrate.

Answer 3

Notice you assume that:

(u+2)/[u^2 + u + 1] = 1/[(u+1/2)^2 + 3/4]

but the denominator is really the same thing, therefore it is the same as saying:
u+2 = 1, that is true when u = -1 and false in all other cases.