# Integration with Partial Fraction Decomposition

## Homework Statement

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

## 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:

$-2x + 4 = A(x-1)^{(2)}(x^{(2)}+1) + B(x-1)(x^{(2)}+1) + (Cx+D)*(x-1)(x-1)^{(2)}$

Once expanded :

$-2x + 4 = Ax^4 - A2x^3 + A2x^2 - A2x + A + Bx^3 -Bx^2 +Bx - B + Cx^4 - C3x^3 + C3x^2 - Cx$
$+Dx^3 - D3x^2 + D3x - D$

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

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SteamKing
Staff Emeritus
Homework Helper
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)$^{2}$(x$^{2}$+1), your partial fraction decomposition should have individual factors
of A/(x$^{2}$+1), B/(x - 1), and C/(x - 1)$^{2}$.

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

After talking with the old teacher/friend I am here:

$-2x + 4 = (A+B+D)x^{3} + (-A +C - 2D + E)x^{2} + (A + B + D -2E)x -A + C + E$

Still working towards the answer. Please let me know if you see any mistakes

Mark44
Mentor
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.

BiGyElLoWhAt
Gold Member

## Homework Statement

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

## 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:

$-2x + 4 = A(x-1)^{(2)}(x^{(2)}+1) + B(x-1)(x^{(2)}+1) + (Cx+D)*(x-1)(x-1)^{(2)}$

Once expanded :

$-2x + 4 = Ax^4 - A2x^3 + A2x^2 - A2x + A + Bx^3 -Bx^2 +Bx - B + Cx^4 - C3x^3 + C3x^2 - Cx$
$+Dx^3 - D3x^2 + D3x - D$

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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