Question

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

Answer 1

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`