# How do you use partial fraction decomposition to decompose the fraction to integrate (6s^3) /(s^(4) - 5s^(2) +4)?

May 22, 2015

The denominator can be factored as:

${s}^{4} - 5 {s}^{2} + 4 = \left({s}^{2} - 4\right) \left({s}^{2} - 1\right) = \left(s - 2\right) \left(s + 2\right) \left(s - 1\right) \left(s + 1\right)$

Hence, we can write $\frac{6 {s}^{3}}{{s}^{4} - 5 {s}^{2} + 4} = \frac{A}{s - 2} + \frac{B}{s + 2} + \frac{C}{s - 1} + \frac{D}{s + 1}$

Now multiply everything by $\left(s - 2\right) \left(s + 2\right) \left(s - 1\right) \left(s + 1\right)$ to get

$6 {s}^{3} = A \left(s + 2\right) \left(s - 1\right) \left(s + 1\right) + B \left(s - 2\right) \left(s - 1\right) \left(s + 1\right) + C \left(s - 2\right) \left(s + 2\right) \left(s + 1\right) + D \left(s - 2\right) \left(s + 2\right) \left(s - 1\right)$

You want this equation to be true for all $s$. This allows you to solve for $A , B , C$, and $D$. The quickest way to do this is to plug in, successively, $s = 1$, $s = - 1$, $s = 2$, and $s = - 2$ (even though the original equation was undefined at these values).

$s = 1 \setminus R i g h t a r r o w 6 = C \setminus \cdot \left(- 1\right) \setminus \cdot 3 \setminus \cdot 2 = - 6 C \setminus R i g h t a r r o w C = - 1$

$s = - 1 \setminus R i g h t a r r o w - 6 = D \setminus \cdot \left(- 3\right) \setminus \cdot 1 \setminus \cdot \left(- 2\right) = 6 D \setminus R i g h t a r r o w D = - 1$

$s = 2 \setminus R i g h t a r r o w 48 = A \setminus \cdot 4 \setminus \cdot 1 \setminus \cdot 3 = 12 A \setminus R i g h t a r r o w A = 4$

$s = - 2 \setminus R i g h t a r r o w - 48 = B \setminus \cdot \left(- 4\right) \setminus \cdot \left(- 3\right) \setminus \cdot \left(- 1\right) = - 12 B \setminus R i g h t a r r o w B = 4$

Hence,

$\setminus \int \setminus \frac{6 {s}^{2}}{{s}^{4} - 5 {s}^{2} + 4} \setminus \mathrm{ds}$

$= \setminus \int \setminus \frac{4}{s - 2} \setminus \mathrm{ds} + \setminus \int \setminus \frac{4}{s + 2} \setminus \mathrm{ds} - \setminus \int \setminus \frac{1}{s - 1} \setminus \mathrm{ds} - \setminus \int \setminus \frac{1}{s + 1} \setminus \mathrm{ds}$

$= 4 \ln | s - 2 | + 4 \ln | s + 2 | - \ln | s - 1 | - \ln | s + 1 | + C$

$= \ln | \setminus \frac{{\left({s}^{2} - 4\right)}^{4}}{{s}^{2} - 1} | + C$