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

Question

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

Emily Ammerman · Accepted Answer

#int1/(s(s-1)^2)ds=lns-ln(s-1)-1/(s-1)# Partial fractions of #1/(s(s-1)^2)hArrA/s+B/(s-1)+C/(s-1)^2# = #(A(s-1)^2+Bs(s-1)+Cs)/(s(s-1)^2)# = #(A(s^2-2s+1)+Bs^2-Bs+Cs)/(s(s-1)^2)# = #(s^2(A+B)+s(-2A-B+C)+A)/(s(s-1)^2)# Hence #A=1#, #A+B=0# and #-2A-B+C=0# or #A=1#, #B=-1# and #C=1# and #1/(s(s-1)^2)=1/s-1/(s-1)+1/(s-1)^2# Hence #int1/(s(s-1)^2)ds=int1/sds-int1/(s-1)ds+int1/(s-1)^2ds# #lns-ln(s-1)-1/(s-1)#

Evelyn Agard · Answer

undefined To integrate $ \int \frac{1}{s(s - 1)^2} $ using partial fractions, we first express the given fraction as a sum of simpler fractions.

Let's denote the given fraction as $ \frac{1}{s(s - 1)^2} $. We aim to express it in the form:

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

where $ A $, $ B $, and $ C $ are constants we need to determine.

To find $ A $, $ B $, and $ C $, we clear the denominators by multiplying both sides of the equation by $ s(s - 1)^2 $:

$$ 1 = A(s - 1)^2 + Bs(s - 1) + Cs $$

Expanding and simplifying:

$$ 1 = As^2 - 2As + A + Bs^2 - Bs + Cs $$

Now, we equate the coefficients of corresponding terms on both sides of the equation:

For $ s^2 $: $ A + B = 0 $
For $ s $: $ -2A - B + C = 0 $
For the constant term: $ A = 1 $

Solving these equations simultaneously will give us the values of $ A $, $ B $, and $ C $.

From the first equation, we get $ A = -B $. Plugging this into the second equation:

$$ -2(-B) - B + C = 0 $$
$$ 2B - B + C = 0 $$
$$ B + C = 0 $$
$$ C = -B $$

Now, substituting $ A = 1 $ and $ C = -B $ into $ A = -B $, we find $ A = -1 $ and $ B = 1 $.

So, the partial fraction decomposition is:

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

Now, we can integrate each term separately.