Partial Fraction Problem

  • Thread starter jdawg
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  • #1
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Homework Statement


∫(5x2+20x+6)/(x3+2x2+x


Homework Equations





The Attempt at a Solution


(5x2+20x+6)/(x3+(x(x2+2x+1)

(5x2+20x+6)=(A/x)+(B/(x+1))+(C/(x+1))

(5x2+20x+6)=x2(A+B+C)+x(2A+B+C)+A

5=A+B+C
20=2A+B+C
6=A

It's not coming out quite right. Did I maybe factor the denominator incorrectly?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


∫(5x2+20x+6)/(x3+2x2+x


Homework Equations





The Attempt at a Solution


(5x2+20x+6)/(x3+(x(x2+2x+1)

(5x2+20x+6)=(A/x)+(B/(x+1))+(C/(x+1))

You want$$
\frac{5x^2+20x+6}{x(x+1)^2}=\frac A x + \frac{Bx+C}{(x+1)^2}$$
 
  • #3
367
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Ohh! But wouldn't you need one for x+1?:

A/x + (Bx+C)/(x+1)2 + (Dx+E)/(x+1)
 
  • #4
Dick
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Ohh! But wouldn't you need one for x+1?:

A/x + (Bx+C)/(x+1)2 + (Dx+E)/(x+1)

No, you only need three independent variables. You could write it as $$\frac A x + \frac{Bx+C}{(x+1)^2}$$ as LCKurtz did, or you could write it as $$\frac A x + \frac{B}{(x+1)^2}+\frac{C}{(x+1)}$$.
 
  • #5
367
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Oh ok! Thanks for clearing that up :)
 

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