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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?

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  • $\begingroup$ 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. $\endgroup$
    – Clayton
    Feb 15, 2018 at 2:19
  • $\begingroup$ 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. $\endgroup$
    – pjs36
    Feb 15, 2018 at 2:40

1 Answer 1

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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.

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

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