urgent help requested

[val1]

please can you help me to resolve this into partial fractions/polynomials?

2 / [(x-1)^2(x^2+1)]

 

[stapel]

Start by forming the various fractional forms:

. . . . .A/(x - 1) + B/(x - 1)² + (Cx + D)/(x² + 1)

Set this equal to your fraction, cross-multiply to clear the denominators, compare coefficients, and solve the resulting system for the values of A, B, C, and D.

Where are you lost in this process? Please reply showing how far you've gotten. Thank you.

Eliz.

 

[val1]

please check my answer

This is what I got:
2 / [(x-1)^2 (x^2+1)] = A/(x - 1) + B/(x - 1)2 + (Cx + D)/(x2 + 1)

= [A(x-1)(x^2+1) + B(x^2+1) + (Cx+D) (x-1)^2 ] / [(x-1)^2 (x^2+1)]

{not sure if the denominator is correct, so unsure if the cross-multiplication is right}

Looking at the numerator, if you make x=1, you get B=1

Then compare coefficients:

Expanding the brackets,

=Ax^3 - Ax^2 +Ax – A + Bx + B + Cx^3 + Cx + Dx^3 +D

=(A+C+D)x^ 3 + Ax^2 + (A+B+C)x – A + B + D

The coefficient of x^2 is 0, so A=0

Coefficient of x is 0, so A+B+C = 0
We already know that A=0 and B=1 so C=-1

Coefficient of x^3 is 0, so A+C+D = 0
Substituting for the known values of A & C leaves us with
D = 1

I’m not sure that this is correct for many reasons including having A=0, but I hope you can see where I’m having problems and can help me solve this correctly.

Thanks

 

[soroban]

Re: please check my answer

Hello, val1!

This is what I got:

2 / [(x-1)²(x²+1)] .= .A/(x - 1) + B/(x - 1)² + (Cx + D)/(x² + 1)

2 .= .A(x-1)(x² + 1) + B(x² + 1) + (Cx + D) (x - 1)²

If you make x=1, you get B=1 . . . . correct to this point

Then compare coefficients . . . . right!

Expanding the brackets . . . . an error here

A(x³ - x² + x - 1) + B(x² + 1) +C(x³ - 2x² + x) + D(x² - 2x + 1)

. . = .Ax³ - Ax² + Ax - A + Bx² + B +Cx³ - 2Cx² + Cx + Dx² - 2Dx + D

. . = .(A + C)x³ + (-A + B - 2C + D)x² + (A + C - 2D)x + (-A + B + D)

And we have: . A . . . + .C . . . .= .0
. . . . . . . . . . . .-A + B - 2C + D .= .0
. . . . . . . . . . . . A . . . + .C - 2D .= .0
. . . . . . . . . . . .-A + B . . . + . D .= .2

I got: .A = -1, B = 1, C = 1, D = 0