Partial Frac. Decomp. Integral ( long div? )

  • Thread starter theRukus
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  • #1
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Homework Statement


\int^1_0 \frac{-3x^3+12x+30}{x^2-x-6}dx


Homework Equations





The Attempt at a Solution


I've attempted long division, but the long division does not seem to come to an end.. I'm not sure what to make of this.. If the long division does not end, does this mean that the polynomial has no factors...? I'm confused, can someone give me a push in the right direction?

Thanks PhysicsForums!
 

Answers and Replies

  • #2
vela
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Homework Statement


[tex]\int^1_0 \frac{-3x^3+12x+30}{x^2-x-6}dx[/tex]


Homework Equations





The Attempt at a Solution


I've attempted long division, but the long division does not seem to come to an end.. I'm not sure what to make of this.. If the long division does not end, does this mean that the polynomial has no factors...? I'm confused, can someone give me a push in the right direction?

Thanks PhysicsForums!
The division ends when the degree of the remainder is less than the degree of the divisor. In this case, when the remainder is of the form Cx+D, you're done. The result of your division will be that
[tex]\frac{-3x^3+12x+30}{x^2-x-6} = Ax + B + \frac{Cx+D}{x^2-x-6}[/tex]
 
  • #3
34,694
6,397

Homework Statement


\int^1_0 \frac{-3x^3+12x+30}{x^2-x-6}dx


Homework Equations





The Attempt at a Solution


I've attempted long division, but the long division does not seem to come to an end.. I'm not sure what to make of this.. If the long division does not end, does this mean that the polynomial has no factors...? I'm confused, can someone give me a push in the right direction?

Thanks PhysicsForums!
The long division does come to an end. You should have gotten -3x - 3 + <remainder>/(x2 - x - 6).

What I'm calling <remainder> is a first-degree polynomial.
 
  • #4
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Got it! Thanks guys, I owe an infinite amount of thanks to PhysicsForums..!
 

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