How do you express $\frac{1}{{(s + 1)}^{2}}$ $1/(s+1)^2$ in partial fractions?

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Answer 1 · 2016-05-07T10:25:41

$\frac{1}{{(s + 1)}^{2}}$ $1/(s+1)^2$ is already in its partial fractions form.

Partial fractions of $\frac{1}{{(s + 1)}^{2}}$ $1/(s+1)^2$ will be of type

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

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

or $\frac{1}{{(s + 1)}^{2}} = \frac{A s + A + B}{{(s + a)}^{2}}$ $1/(s+1)^2=(As+A+B)/(s+a)^2$

Equating coefficients of numerator, we have $A = 0$ $A=0$ and $A + B = 1$ $A+B=1$ or $B = 1$ $B=1$ .

Hence $\frac{1}{{(s + 1)}^{2}} = \frac{0}{s + 1} + \frac{1}{{(s + 1)}^{2}}$ $1/(s+1)^2=0/(s+1)+1/(s+1)^2$

or $\frac{1}{{(s + 1)}^{2}} = \frac{1}{{(s + 1)}^{2}}$ $1/(s+1)^2=1/(s+1)^2$

It is apparent that $\frac{1}{{(s + 1)}^{2}}$ $1/(s+1)^2$ is already in its partial fractions form.

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