# Integration dont think this is what im meant to do

1. Jan 19, 2009

### Dell

integration,,, dont think this is what im meant to do

in this integration problem, looks simple, but i must be missing something

$$\int$$$$\frac{x-5}{x2-2x+2}$$

how do i deal with a propper fraction where i cannot simplify the denominator

2. Jan 19, 2009

### MathematicalPhysicist

Re: integration,,, dont think this is what im meant to do

(x-5)/(x^2-2x+2)
The roots of x^2-2x+2 are x1,2=1+-sqrt(4-8)=1+-2i
i.e (x-5)/((x-1+2i)(x-1-2i))=A/(x-1+2i)+B/(x-1-2i)

Do you want me to stop here, or do you need another advice on how to procceed?

3. Jan 19, 2009

### HallsofIvy

Staff Emeritus
Re: integration,,, dont think this is what im meant to do

I'm not at all clear what loop quantum gravity is trying to do!

Completing the square in the denominator gives $x^2+ 2x+ 1 +1=(x+1)^2+ 1$
Substitute u= x+ 2 so x= u- 2 and du= dx, x- 5= u- 7. The integral becomes
$$\int\frac{u- 7}{u^2+1}du= \int \frac{u}{u^2+1}du+ \int \frac{1}{u^2+1}du$$
The first can be done with the second substitution $v= x^2+ 1$ and the second is an arctangent.

4. Jan 19, 2009

### MathematicalPhysicist

Re: integration,,, dont think this is what im meant to do

I am using partial fraction decomposition here, i.e there'a a thereoem that
if P(x)/Q(x) and degQ(x)>=deg(P(x)), and Q(x)=(x-x0)^m1*....(x-xn)^m_n, then you can write it as: P(x)/Q(x)=A1/(x-x0)^m1+....+An/(x-xn)^m_n, are my intentions now clearer?