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Integration with Partial Fraction Decomposition

  • Thread starter m0gh
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  • #1
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Homework Statement


[itex]\int \frac{-2x + 4}{(x-1)^{(2)}(x^{(2)}+1)}[/itex]


Homework Equations





The Attempt at a Solution



I've done the problem a couple times but the answers keep coming out differently so I'm assuming I am messing up the setup.

This is what I have for the first part of the setup:

[itex] -2x + 4 = A(x-1)^{(2)}(x^{(2)}+1) + B(x-1)(x^{(2)}+1) + (Cx+D)*(x-1)(x-1)^{(2)} [/itex]

Once expanded :

[itex] -2x + 4 = Ax^4 - A2x^3 + A2x^2 - A2x + A + Bx^3 -Bx^2 +Bx - B + Cx^4 - C3x^3 + C3x^2 - Cx[/itex]
[itex] +Dx^3 - D3x^2 + D3x - D [/itex]


Can anyone let me know if I'm right up to this point?
 
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Answers and Replies

  • #2
SteamKing
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It's not clear why some exponents are enclosed with parentheses () while others are not.

Given that the denominator of the original rational expression is (x - 1)[itex]^{2}[/itex](x[itex]^{2}[/itex]+1), your partial fraction decomposition should have individual factors
of A/(x[itex]^{2}[/itex]+1), B/(x - 1), and C/(x - 1)[itex]^{2}[/itex].
 
  • #3
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Just talked to an old teacher/friend and she said it should be set up as A/x-1 + (Bx+C)/(x-1)^(2) + (Cx + D)/(x^(2) + 1) so I'm not sure what you mean, Steam King
 
  • #4
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After talking with the old teacher/friend I am here:

[itex] -2x + 4 = (A+B+D)x^{3} + (-A +C - 2D + E)x^{2} + (A + B + D -2E)x -A + C + E [/itex]

Still working towards the answer. Please let me know if you see any mistakes
 
  • #5
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Just talked to an old teacher/friend and she said it should be set up as A/x-1 + (Bx+C)/(x-1)^(2) + (Cx + D)/(x^(2) + 1) so I'm not sure what you mean, Steam King
No, this isn't correct. The second term should be B/(x - 1)2.

Also, you are missing some needed parentheses in your first term, which should be A/(x - 1), when written on a single line.
 
  • #6
BiGyElLoWhAt
Gold Member
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Homework Statement


[itex]\int \frac{-2x + 4}{(x-1)^{(2)}(x^{(2)}+1)}[/itex]


Homework Equations





The Attempt at a Solution



I've done the problem a couple times but the answers keep coming out differently so I'm assuming I am messing up the setup.

This is what I have for the first part of the setup:

[itex] -2x + 4 = A(x-1)^{(2)}(x^{(2)}+1) + B(x-1)(x^{(2)}+1) + (Cx+D)*(x-1)(x-1)^{(2)} [/itex]

Once expanded :

[itex] -2x + 4 = Ax^4 - A2x^3 + A2x^2 - A2x + A + Bx^3 -Bx^2 +Bx - B + Cx^4 - C3x^3 + C3x^2 - Cx[/itex]
[itex] +Dx^3 - D3x^2 + D3x - D [/itex]


Can anyone let me know if I'm right up to this point?
Setup: ##\frac{-2x+4}{(x-1)^2(x^2+1)}## and work that out, granting:
##\frac{-2x+4}{(x-1)^2(x^2+1)}=\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}##
now multiply both sides by ## (x-1)^2(x^2+1) ## which grants
##-2x+4 = A\frac{(x-1)}{(x-1)}(x-1)(x^2+1) + B\frac{(x-1)^2}{(x-1)^2 }(x^2+1) +(Cx+D) \frac{(x^2+1)}{ (x^2+1) }(x-1)^2 ##
I don't know whether what you did was right or not, but it looks as though you didn't cancel your same terms (judging by the ##x^4##) but do this and you should end up with the right answer.

Also (for future reference)
http://tutorial.math.lamar.edu/Classes/Alg/PartialFractions.aspx
That has a nice little reference sheet on it.
 
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