Integration dont think this is what im meant to do

[Dell]

integration,,, don't think this is what I am meant to do

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

$$\int$$[tex]\frac{x-5}{x²-2x+2}[/tex]

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

 

[MathematicalPhysicist]

(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?

 

[HallsofIvy]

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.

 

[MathematicalPhysicist]

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?