Integration with Partial Fraction Decomposition

[m0gh]

Homework Statement

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

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:

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

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?

 

[SteamKing]

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)$^{2}$.

 

[m0gh]

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

 

[m0gh]

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]

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]

Homework Statement

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

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:

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

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.