Use elementary fractions to write the fraction 1/(x^3+x^2+x+1)

To use in elementary fractions to write the fraction
1/(x^3+x^2+x+1).

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

Therefore  let 1/(x^2+x^2+x+1) = a/(x+1) +
(bx+c)/(x^2+1)....(1).

 We have to determine a, b and
c.

We multiply both sides of (1) by (x+1)(x^2+1) and we
get:

1 = a(x^2+1) +
(bx+c)(x+1)

 1 =  ax^2+a
+bx^2+bx+cx+c.

1 = (a+b)x^2+(b+c)x+ (a+c). We treat this as
an identity. So we can equate the coefficient of like terms on both
sides:

x^2 terms: 0 =
a+b...(1).

x terms = 0 =
b+c......(2).

Constant terms: 1 = a+c, or a+c =
1.....(3)

 (1) -(2): a - c =
0...(4).

(3)+(4) : 2a = 1. So a=
1/2.

(3-(4): 2c =  1. So c =
1/2.

Therfore we put a = 1/2 in (1) and get a+b = 0, or
(1/2 +b = 0. S = b= -1/2.

Therefore a = 1/2, b=-1/2 and c =
1/2.

So

1/(x^3+x^2+x+1) =
1/{2(x+1)} + (-x+1)/{2(x^2+1)}

1/(x^3+x^2+x+1) = 1/(2(x+1)}
+1/{2{x^2+1) - x/{2(x^2+1)}.