How do you write the partial fraction decomposition of the rational expression #[s(s+6)] / [(s+3)(s^2+6s+18)]#?

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Answer 1 · 2017-03-03T08:06:12

#[s(s+6)] / [(s+3)(s^2+6s+18)]=2/(3(s+3))+(s-12)/(3(s^2+6s+18))#

Let the partial fractions be

#[s(s+6)] / [(s+3)(s^2+6s+18)]=A/(s+3)+(Bs+C)/(s^2+6s+18)#

or #[s^2+6s] / [(s+3)(s^2+6s+18)]=[A(s^2+6s+18)+(Bs+C)(s+3)]/[(s+3)(s^2+6s+18)]#

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

comparing coefficients of #s^2#, #s# and constant term, we get

#A+B=1#, #6A+C+3B=1# and #18A+3C=0# or #6A+C=0#

Hence #3B=1# or #B=1/3#,

then #A=1-1/3=2/3# and then #C=--6xx2/3=-4#

and hence

#[s(s+6)] / [(s+3)(s^2+6s+18)]=(2/3)/(s+3)+(s/3-4)/(s^2+6s+18)#

#[s(s+6)] / [(s+3)(s^2+6s+18)]=2/(3(s+3))+(s-12)/(3(s^2+6s+18))#