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

Jun 25, 2016

Split apart the fraction using typical decomposition rules:

$\frac{1}{{s}^{2} \left(3 s + 5\right)} = \frac{A}{s} + \frac{B}{s} ^ 2 + \frac{C}{3 s + 5}$

Multiply through by ${s}^{2} \left(3 s + 5\right)$:

$1 = A s \left(3 s + 5\right) + B \left(3 s + 5\right) + C {s}^{2}$

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

$1 = 5 B$

$B = \frac{1}{5}$

Let $s = - \frac{5}{3}$, making both the $A$ and $B$ terms become $0$:

$1 = C {\left(- \frac{5}{3}\right)}^{2} = C \left(\frac{25}{9}\right)$

$C = \frac{9}{25}$

Arbitrarily let $s = 5$ to solve for $A$, using $B = \frac{1}{5}$ and $C = \frac{9}{25}$:

$1 = A \left(5\right) \left(3 \cdot 5 + 5\right) + \frac{1}{5} \left(3 \cdot 5 + 5\right) + \frac{9}{25} {\left(5\right)}^{2}$

$1 = A \left(100\right) + 4 + 9$

$- 12 = 100 A$

$A = - \frac{3}{25}$

Thus:

$\frac{1}{{s}^{2} \left(3 s + 5\right)} = - \frac{3}{25 s} + \frac{1}{5 {s}^{2}} + \frac{9}{25 \left(3 s + 5\right)}$

So:

$\int \frac{1}{{s}^{2} \left(3 s + 5\right)} \mathrm{ds} = - \frac{3}{25} \int \frac{1}{s} \mathrm{ds} + \frac{1}{5} \int {s}^{-} 2 \mathrm{ds} + \frac{9}{25} \int \frac{1}{3 s + 5} \mathrm{ds}$

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

$= - \frac{3}{25} \ln \left(\left\mid s \right\mid\right) - \frac{1}{5 s} + \frac{3}{25} \ln \left(\left\mid 3 s + 5 \right\mid\right) + C$