# How do you integrate int [s(s+6)] / [(s+3)(s^2+6s+18)] using partial fractions?

Dec 19, 2016

The answer is ==-ln(∣s+3∣)+ln(∣s^2+6s+18∣)+C

#### Explanation:

Let's do the decomposition into partial fractions

$\frac{s \left(s + 6\right)}{\left(s + 3\right) \left({s}^{2} + 6 s + 18\right)} = \frac{A}{s + 3} + \frac{B s + C}{{s}^{2} + 6 s + 18}$

$= \frac{A \left({s}^{2} + 6 s + 18\right) + \left(B s + C\right) \left(x + 3\right)}{\left(s + 3\right) \left({s}^{2} + 6 s + 18\right)}$

So

$s \left(s + 6\right) = A \left({s}^{2} + 6 s + 18\right) + \left(B s + C\right) \left(s + 3\right)$

Let $s = 0$, $\implies$, $0 = 18 A + 3 C$, $\implies$, $C = - 6 A$

Let $s = - 3$, $\implies$, $- 9 = 9 A$, $\implies$, $A = - 1$

Coefficients of ${s}^{2}$, $\implies$, $1 = A + B$, $\implies$, $B = 2$

Therefore,

$\frac{s \left(s + 6\right)}{\left(s + 3\right) \left({s}^{2} + 6 s + 18\right)} = - \frac{1}{s + 3} + \frac{2 s + 6}{{s}^{2} + 6 s + 18}$

$\int \frac{s \left(s + 6\right) \mathrm{ds}}{\left(s + 3\right) \left({s}^{2} + 6 s + 18\right)} = \int - \frac{\mathrm{ds}}{s + 3} + \int \frac{\left(2 s + 6\right) \mathrm{ds}}{{s}^{2} + 6 s + 18}$

=-ln(∣s+3∣)+ln(∣s^2+6s+18∣)+C