Partial Fractions

**coobe** · June 11th 2009, 08:39 AM

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

**alexmahone** · June 11th 2009, 09:28 AM

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

**AMI** · June 11th 2009, 09:32 AM

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..

**coobe** · June 12th 2009, 02:43 AM

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

**HallsofIvy** · June 12th 2009, 02:59 AM

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 $x^2+ 1$ is because it can't be factored!

**AMI** · June 12th 2009, 11:33 AM

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..

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