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

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

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Answer 1 · 2016-06-25T02:05:50

Split apart the fraction using typical decomposition rules:

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

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

$1 = A s (3 s + 5) + B (3 s + 5) + C s^{2}$ $1=As(3s+5)+B(3s+5)+Cs^2$

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

$1 = 5 B$ $1=5B$

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

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

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

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

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

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

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

$- 12 = 100 A$ $-12=100A$

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

Thus:

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

So:

$\int \frac{1}{s^{2} (3 s + 5)} d s = - \frac{3}{25} \int \frac{1}{s} d s + \frac{1}{5} \int s^{- 2} d s + \frac{9}{25} \int \frac{1}{3 s + 5} d s$ $int1/(s^2(3s+5))ds=-3/25int1/sds+1/5ints^-2ds+9/25int1/(3s+5)ds$

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

$= - \frac{3}{25} ln (| s |) - \frac{1}{5 s} + \frac{3}{25} ln (| 3 s + 5 |) + C$ $=-3/25ln(abss)-1/(5s)+3/25ln(abs(3s+5))+C$

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