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

Sep 8, 2016

$\int \frac{1}{s {\left(s - 1\right)}^{2}} \mathrm{ds} = \ln s - \ln \left(s - 1\right) - \frac{1}{s - 1}$

#### Explanation:

Partial fractions of $\frac{1}{s {\left(s - 1\right)}^{2}} \Leftrightarrow \frac{A}{s} + \frac{B}{s - 1} + \frac{C}{s - 1} ^ 2$

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

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

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

Hence $A = 1$, $A + B = 0$ and $- 2 A - B + C = 0$

or $A = 1$, $B = - 1$ and $C = 1$ and

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

Hence $\int \frac{1}{s {\left(s - 1\right)}^{2}} \mathrm{ds} = \int \frac{1}{s} \mathrm{ds} - \int \frac{1}{s - 1} \mathrm{ds} + \int \frac{1}{s - 1} ^ 2 \mathrm{ds}$

$\ln s - \ln \left(s - 1\right) - \frac{1}{s - 1}$