integration by partial fraction?

Question

integrate : [(s^4 + 81) / s(s^2 +9)^2] ds

I know the set up for this question should be

A/s + (Bs +C/ s^2 + 9) + (Ds + E / (S^2 +9)^2)

How do I find the value of A,B,C,D,E?
My teacher said that this question also involves u substitution, but I don't know where.

Answer 1

(s^4 + 81) / [s(s^2 + 9)^2]
= A/s + (Bs + C)/(s^2 + 9) + (Ds + E) / (S^2 + 9)^2

Clear the denominators by multiplying thru by s(s^2 + 9)^2

s^4 + 81
= A(s^2 + 9)^2 + (Bs +C)s(s^2 + 9) + (Ds + E)s

Let's rewrite this expression

s^4 + 0s² + 0s² + 0s + 81
= A(s^4 + 18s² + 81) + (Bs +C)(s³ + 9s) + (Ds + E)s

s^4 + 0s² + 0s² + 0s + 81
= As^4 + 18As² + 81A + Bs^4 + 9Bs² + Cs³ + 9Cs + Ds² + Es

s^4 + 0s² + 0s² + 0s + 81
= (A + B)s^4 + Cs³ + (18A + 9B + D)s² + (9C + E)s + 81A

Now equate the coefficients of each power of s.

1s^4 = (A + B)s^4
0s³ = Cs³
0s² = (18A + 9B + D)s²
0s = (9C + E)s
81 = 81A

Divide by the powers of s and you have five equations in five unknowns.

1 = A + B
0 = C
0 = 18A + 9B + D
0 = 9C + E
81 = 81A
__________

81 = 81A
A = 1

1 = A + B
B = 1 - A = 1 - 1 = 0

C = 0

0 = 9C + E
E = -9C = -9*0 = 0

0 = 18A + 9B + D
0 = 18*1 + 9*0 + D
0 = 18 + D
D = -18

{A,B,C,D,E} = {1,0,0,-18,0}
____________________

(s^4 + 81) / [s(s^2 + 9)^2]
= A/s + (Bs + C)/(s^2 + 9) + (Ds + E) / (S^2 + 9)^2
= 1/s + (0s + 0)/(s^2 + 9) + (-18s + 0) / (S^2 + 9)^2
= 1/s - 18s/(s² + 9)²

This expression can now be integrated.

∫{(s^4 + 81)/[s(s² + 9)²]}ds
= ∫{1/s - 18s/(s² + 9)²}ds
= ∫(ds/s) - 9∫{2s/(s² + 9)²}ds
= ln(s) - 9∫{2s/(s² + 9)²}ds + C
Let
u = s² + 9
du = 2sds

= ln(s) - 9∫{1/u²}du + C
= ln(s) + 9/u + C
= ln(s) + 9/(s² + 9) + C