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#