# Partial Fractions

• Jun 11th 2009, 08:39 AM
coobe
Partial Fractions
hi again!

i have to make partialfractions out of the following integral:

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

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

then polynomial division gives me

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

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

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
• Jun 11th 2009, 09:28 AM
alexmahone
$\frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}$
Multiply throughout by (x-2)
$\frac {1-x}{x^2+1} = A + B\frac{x-2}{x^2+1}$
Sub x=2
$-\frac{1}{5}=A$
$A=-\frac{1}{5}$
• Jun 11th 2009, 09:32 AM
AMI
Quote:

Originally Posted by coobe
$\rightarrow \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}$

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

Did you look here: Partial fraction - Wikipedia, the free encyclopedia ?! It seems pretty well written..
• Jun 12th 2009, 02:43 AM
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 :(
• Jun 12th 2009, 02:59 AM
HallsofIvy
Quote:

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 :http://mathhelpforum.com/calculus/92565-partial-fractions.html#post328806\" rel=\"nofollow\">
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 :(

(Wondering) 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..