# How do you integrate int (2s)/[(s+4)(s-1)] using partial fractions?

May 30, 2016

$\frac{8}{5} \ln \left(\left\mid s + 4 \right\mid\right) + \frac{2}{5} \ln \left(\left\mid s - 1 \right\mid\right) + C$

#### Explanation:

Split up $\frac{2 s}{\left(s + 4\right) \left(s - 1\right)}$ into its partial fraction decomposition:

$\frac{2 s}{\left(s + 4\right) \left(s - 1\right)} = \frac{A}{s + 4} + \frac{B}{s - 1}$

$2 s = A \left(s - 1\right) + B \left(s + 4\right)$

Letting $s = 1$:

$2 \left(1\right) = A \left(1 - 1\right) + B \left(1 + 4\right)$

$2 = 5 B$

$B = \frac{2}{5}$

Letting $s = - 4$:

$2 \left(- 4\right) = A \left(- 4 - 1\right) + B \left(- 4 + 4\right)$

$- 8 = - 5 A$

$A = \frac{8}{5}$

Thus,

$\frac{2 s}{\left(s + 4\right) \left(s - 1\right)} = \frac{8}{5} \left(\frac{1}{s + 4}\right) + \frac{2}{5} \left(\frac{1}{s - 1}\right)$

Splitting up the integral through addition:

$\int \frac{2 s}{\left(s + 4\right) \left(s - 1\right)} \mathrm{ds} = \frac{8}{5} \int \frac{1}{s + 4} \mathrm{ds} + \frac{2}{5} \int \frac{1}{s - 1} \mathrm{ds}$

Both of these are simply integrated through the natural logarithm:

$= \frac{8}{5} \ln \left(\left\mid s + 4 \right\mid\right) + \frac{2}{5} \ln \left(\left\mid s - 1 \right\mid\right) + C$