# Integration by Partial Fractions, Complex Long Division

Compute the following integral

$\int _{-1}^0\frac{x^3-4x+1}{x^2-3x+2}\:dx$

Since the degree of the numerator is greater than that of the denominator, I need to perform long division. However, I am not entirely sure on how to do this when both polynomials are fully extended. I know the factorization of the denominator is $(x-1)(x-2)$ but I am not entirely sure where/if this could help. I think there is a step I am missing.

Any help?

• What confuses you? You just need to use regular long division until the degree of the numerator is less than the denominator. This should be fairly straightforward. Feb 15, 2018 at 2:19
• Since the denominator factors, you can also divide first by $x - 2$, then by $x - 1$ (or vice versa). And for these, you could use synthetic division, which might be a bit quicker than long division. Feb 15, 2018 at 2:40

To divide the numerator. Subtract the largest multiple of the denominator from the numerator that will kill the highest degree term. Repeat as necessary.

$x^3 - 4x + 1 \\ x(x^2 - 3x +2) - x^3 + 3x^2 - 2x + x^3 - 4x + 1\\ x(x^2 - 3x +2) +3x^2 - 6x + 1\\ x(x^2 - 3x +2) +3(x^2-3x + 2) - 3x^2 +9x - 6 +3x^2 - 6x + 1\\ x(x^2 - 3x +2) +3(x^2-3x + 2) +3x -5\\ (x+3)(x^2 - 3x +2) +3x -5\\ \frac {x^3 - 4x + 1}{x^2-3x+2} = x+3 +\frac {3x-5}{x^2-3x+2}$

And then you will need to use partial fractions to do the rest.

$x+3 +\frac {3x-5}{x^2-3x+2} = x+ 3 +\frac {A}{x-2} + \frac B{x-1}\\ x+ 3 +\frac {1}{x-2} + \frac 2{x-1}$

altnernative.

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

Divide the numerator by just one factor at a time.

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