Partial fractions

Question

Answer 1

The expression

- { 1 / [(s^2 + 1)(s + 1)^2] }

Consider

1 / [(s^2 + 1)(s + 1)^2]   =   (As + B) / (s^2 + 1) + C / (s + 1) + D / (s + 1)^2 ---------> (T)

Take the LCM

1 / [(s^2 + 1)(s + 1)^2]   =   [(As + B)(s + 1)^2] / (s^2 + 1) + [C(s^2 + 1)(s + 1)]/ (s + 1) + [D(s^2 + 1)] / (s + 1)^2

1 / [(s^2 + 1)(s + 1)^2]   =   { [(As + B)(s^2 + 1 + 2s)] + [C(s^3 + s^2 + s + 1)] + [Ds^2 + D)] } / [(s^2 + 1)(s + 1)^2]

1   =   As^3 + 2As^2 + As + Bs^2 + 2Bs + B + Cs^3 + Cs^2 + Cs + C + Ds^2 + D

0s^3 + 0s^2 + 0S + 1   =   (A+C)s^3 + (2A+B +C+D)s^2 + (A + 2B + C)s + (B + C + D)

Compare the coefficients o s^3, s^2, s and constant terms

A + C   =   0  --------------------------------->  (1)

2A + B + C + D   =   0  --------------------------------->  (2)

A + 2B + C   =   0  --------------------------------->  (3)

B + C + D   =   1  --------------------------------->  (4)

Substitute value of Eq (4) in Eq (1)

2A + 1   =   0

2A   =   -1

A   =   -1/2

Substitute A = -1/2 in Eq (1)

-1/2 + C   =   0

C  =  1/2

Substitute A = -1/2 and C = 1/2 in Eq (3)

-1/2 + 2B + 1/2   =   0

2B  =   0

B  =  0

Substitute A = -1/2 and C = 1/2 in Eq (4)

0 + 1/2 + D   =   1

D   =   1 - 1/2

D  =  1/2

Substitute A = -1/2, B = 0, C = 1/2 and D = 1/2 in Eq (T)

1 / [(s^2 + 1)(s + 1)^2]   =   (-s/2 + 0) / (s^2 + 1) + (1/2) / (s + 1) + (1/2) / (s + 1)^2

1 / [(s^2 + 1)(s + 1)^2]   =   -s/ 2(s^2 + 1) + 1 / 2(s + 1) + 1 /2 (s + 1)^2

Hence,

- { 1 / [(s^2 + 1)(s + 1)^2] }  =   s/ 2(s^2 + 1) - 1 / 2(s + 1) - 1 /2 (s + 1)^2