please solve this problem ? resolve the partial fraction.?

Question

x^2+x+1
--------------------------------
( x^2 - 1) ( x^2 - 5x+6)

Answer 1

(x^2 + x + 1) / [(x^2 - 1) (x^2 - 5x + 6)]

First thing to do is factor the denominator.

(x^2 + x + 1) / [(x - 1)(x + 1) (x - 2) (x - 3)]

The partial fraction decomposition is


(x^2 + x + 1) / [(x - 1)(x + 1) (x - 2) (x - 3)] =

A/(x - 1) + B/(x + 1) + C/(x - 2) + D/(x - 3)

In order to solve for A, B, C and D, we multiply both sides by
(x - 1)(x + 1)(x - 2)(x - 3), getting

x^2 + x + 1 = A(x + 1)(x - 2)(x - 3) + B(x - 1)(x - 2)(x - 3) +
C(x - 1)(x + 1)(x - 3) + D(x - 1)(x + 1)(x - 2)

Note that this equation is true _for all x_. That means we can assign x ANY value and get our values for A, B, C, and D.

Let x = -1: Notice how we're going to have 0 in the first, third, and fourth expressions on the right hand side (since we have an x + 1 term, and that becomes 0 when x = -1).

(-1)^2 + (-1) + 1 = 0 + B(-1 - 1)(-1 - 2)(-1 - 3) + 0 + 0
1 = B(-2)(-3)(-4)
1 = B(-24), so B = -1/24.

Let x = 1. Then

1^2 + 1 + 1 = A(1 + 1)(1 - 2)(1 - 3) + 0 + 0 + 0
3 = A(2)(-1)(-2)
3 = A(4), so A = 3/4

Let x = 2. Then

2^2 + 2 + 1 = 0 + 0 + C(2 - 1)(2 + 1)(2 - 3) + 0
4 + 2 + 1 = C(1)(3)(-1)
7 = C(-3), so C = -7/3

Let x = 3. Then

3^2 + 3 + 1 = 0 + 0 + 0 + D(3 - 1)(3 + 1)(3 - 2)
9 + 3 + 1 = D(2)(4)(1)
13 = D(8), therefore D = 13/8

So A = 3/4, B = (-1/24), C = (-7/3), D = 13/8. Recall that

(x^2 + x + 1) / [(x - 1)(x + 1) (x - 2) (x - 3)] =
A/(x - 1) + B/(x + 1) + C/(x - 2) + D/(x - 3)

So our partial fraction decomposition is

(x^2 + x + 1) / [(x - 1)(x + 1) (x - 2) (x - 3)] =
(3/4) [1/(x - 1)] - (1/24) [1/(x + 1)] - (7/3) [1/(x - 2)] +
(13/8)[1/(x - 3)]

Answer 2

the denominator is (x-1)(x+1)(x-2)(x-3) so you write

(X^2 +x+1) /(x^2-1)(x^2-5x+6) =a/(x-1) +b/(x+1) +c/(x-2) +d/(x-3)

reducing to common denominator the numerators must be identical

x^2+x+1 = a(x+1)(x-2)(x-3)+b(x-1)(x-2)(x-3)+c(x-1)(x+1)(x-3) +
d(x-1)(x+1)(x-2)
As both sides must be identical polinoms the take the sam value for every x

So put x= 1 ==> 3= 4a so a=3/4
put x0-1 1=-24b so b= -1/24
put x=2 7=-3c soc=-7/3
put x=3 13=8d so d= 13/8

So put these numbers in your partial fractions

Answer 3

x^2+x+1
------------------------------ =
( x^2 - 1) ( x^2 - 5x+6)

x^2+x+1
------------------------------ =
(x + 1)(x - 1)(x - 2)(x - 3)

A/(x + 1) + B/(x - 1) + C/(x - 2) + D/(x - 3)
A = (1 - 1 + 1)/((-1 - 1)(-1 - 2)(-1 - 3)) = -1/24
B = (1 + 1 + 1)/((1 + 1)(1 - 2)(1 - 3)) = 3/4
C = (4 + 2 + 1)/((2 + 1)(2 - 1)(2 - 3)) = -7/3
D = (9 + 3 + 1)/((3 + 1)(3 - 1)(3 - 2)) = 13/8

-1/(24(x + 1)) + 3/(4(x - 1)) - 7/(3(x - 2)) + 13/(8(x - 3)) =
(1/24)[-1/(x + 1) + 18/(x - 1) - 56/(x - 2) + 39/(x - 3)]

Answer 4

2/(x+1)-3x+1/x