How do you integrate $int [s(s+6)] / [(s+3)(s^2+6s+18)]$ using partial fractions?

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Answer 1 · 2016-12-19T13:12:57

The answer is $==-ln(∣s+3∣)+ln(∣s^2+6s+18∣)+C$

Let's do the decomposition into partial fractions

$(s(s+6))/((s+3)(s^2+6s+18))=A/(s+3)+(Bs+C)/(s^2+6s+18)$

$=(A(s^2+6s+18)+(Bs+C)(x+3))/((s+3)(s^2+6s+18))$

So

$s(s+6)=A(s^2+6s+18)+(Bs+C)(s+3)$

Let $s=0$ , $=>$ , $0=18A+3C$ , $=>$ , $C=-6A$

Let $s=-3$ , $=>$ , $-9=9A$ , $=>$ , $A=-1$

Coefficients of $s^2$ , $=>$ , $1=A+B$ , $=>$ , $B=2$

Therefore,

$(s(s+6))/((s+3)(s^2+6s+18))=-1/(s+3)+(2s+6)/(s^2+6s+18)$

$int(s(s+6)ds)/((s+3)(s^2+6s+18))=int-(ds)/(s+3)+int((2s+6)ds)/(s^2+6s+18)$

$=-ln(∣s+3∣)+ln(∣s^2+6s+18∣)+C$

How do you integrate int [s(s+6)] / [(s+3)(s^2+6s+18)]int [s(s+6)] / [(s+3)(s^2+6s+18)] using partial fractions?