1. ## Partial Fractions

hi again!

i have to make partialfractions out of the following integral:

$\displaystyle \int \frac{1-x}{x^3-2x^2+x-2} dx$

guessing a critical point for the denominator gives me $\displaystyle x_{0}=2$

then polynomial division gives me

$\displaystyle (x^3-2x^2+x-2) / (x-2) = x^2+1$

$\displaystyle \rightarrow \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}$

is this correct ? if so, i dont seem to get an equation system that is solvable...

could you give me a quick hint to do partial fractions in general ? like the different cases etc..., my professors script is a mess
thanks

2. $\displaystyle \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}$
Multiply throughout by (x-2)
$\displaystyle \frac {1-x}{x^2+1} = A + B\frac{x-2}{x^2+1}$
Sub x=2
$\displaystyle -\frac{1}{5}=A$
$\displaystyle A=-\frac{1}{5}$

3. Originally Posted by coobe
$\displaystyle \rightarrow \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}$
Actually, it is $\displaystyle \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{Bx+C}{x^2+1}$
$\displaystyle \Rightarrow 1-x=A(x^2+1)+(Bx+C)(x-2)$ $\displaystyle \Rightarrow A+B=0,-2B+C=-1,A-2C=1$ $\displaystyle \Rightarrow A=-\frac{1}{5},B=\frac{1}{5},C=-\frac{3}{5}$.

Did you look here: Partial fraction - Wikipedia, the free encyclopedia ?! It seems pretty well written..

4. sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither

5. Originally Posted by coobe
sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither
What do you mean by "two factors in the denominator"? The reason we need Bx+ C over $\displaystyle x^2+ 1$ is because it can't be factored!

6. Originally Posted by coobe
sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither
Did you look at the link?? It's the case where you have an irreducible second degree factor in the denominator, just like HallsofIvy said..