How do you integrate $1/[s^2(3s+5)]$ using partial fractions?

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Answer 1 · 2016-06-25T02:05:50

Split apart the fraction using typical decomposition rules:

$1/(s^2(3s+5))=A/s+B/s^2+C/(3s+5)$

Multiply through by $s^2(3s+5)$ :

$1=As(3s+5)+B(3s+5)+Cs^2$

Let $s=0$ , making both the $A$ and $C$ terms become $0$ :

$1=5B$

$B=1/5$

Let $s=-5/3$ , making both the $A$ and $B$ terms become $0$ :

$1=C(-5/3)^2=C(25/9)$

$C=9/25$

Arbitrarily let $s=5$ to solve for $A$ , using $B=1/5$ and $C=9/25$ :

$1=A(5)(3*5+5)+1/5(3*5+5)+9/25(5)^2$

$1=A(100)+4+9$

$-12=100A$

$A=-3/25$

Thus:

$1/(s^2(3s+5))=-3/(25s)+1/(5s^2)+9/(25(3s+5))$

So:

$int1/(s^2(3s+5))ds=-3/25int1/sds+1/5ints^-2ds+9/25int1/(3s+5)ds$

Using typical integration rules (don't forget to substitute in the final integral):

$=-3/25ln(abss)-1/(5s)+3/25ln(abs(3s+5))+C$

How do you integrate 1/[s^2(3s+5)]1/[s^2(3s+5)] using partial fractions?