How do you integrate ∫s(s+6)(s+3)(s2+6s+18) using partial fractions?

1 Answer
Narad T.
Dec 19, 2016

The answer is ==−ln(∣s+3∣)+ln(∣s2+6s+18∣)+C

Explanation:

Let's do the decomposition into partial fractions

s(s+6)(s+3)(s2+6s+18)=As+3+Bs+Cs2+6s+18

=A(s2+6s+18)+(Bs+C)(x+3)(s+3)(s2+6s+18)

So

s(s+6)=A(s2+6s+18)+(Bs+C)(s+3)

Let s=0, ⇒, 0=18A+3C, ⇒, C=−6A

Let s=−3, ⇒, −9=9A, ⇒, A=−1

Coefficients of s2, ⇒, 1=A+B, ⇒, B=2

Therefore,

s(s+6)(s+3)(s2+6s+18)=−1s+3+2s+6s2+6s+18

∫s(s+6)ds(s+3)(s2+6s+18)=∫−dss+3+∫(2s+6)dss2+6s+18

=−ln(∣s+3∣)+ln(∣s2+6s+18∣)+C

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