Polynomial long division

[euonym]

Hi,

I need to find the anti-derivative of this function:

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

My problem lies in how to approach the polynomial division. Any ideas?

 

[galactus]

First, if you factor the top and bottom completely, an x-1 cancels and you're left with:

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

Add and subtract $\displaystyle x^{2}-3x+2$:

$\displaystyle \frac{x^{2}+x-1-(x^{2}-3x+2)+(x^{2}-3x+2)}{x^{2}-3x+2}=\frac{x^{2}+x-1-(x^{2}-3x+2)}{x^{2}-3x+2}+\frac{x^{2}-3x+2}{x^{2}-3x+2}=\frac{4x-3}{(x-2)(x-1)}+1$

Now, you can use partial fractions.

$\displaystyle \frac{A}{x-2}+\frac{B}{x-1}=4x-3$

OR do it without partial fractions:

add and subtract x-2:

$\displaystyle \frac{4x-3+(x-2)-(x-2)}{(x-2)(x-1)}=\frac{5(x-1)-(x-2)}{(x-2)(x-1)}=\frac{5}{x-2}-\frac{1}{x-1}$

Don't forget the 1. It can be integrated separately.

 

[euonym]

Thanks for the help