How do you integrate #1/[s^2(3s+5)]# using partial fractions?

Question

Scarlett Allison · Accepted Answer

Split apart the fraction using typical decomposition rules:

#1/(s^2(3s+5))=A/s+B/s^2+C/(3s+5)#

Multiply through by #s^2(3s+5)#:

#1=As(3s+5)+B(3s+5)+Cs^2#

Let #s=0#, making both the #A# and #C# terms become #0#:

#1=5B#

#B=1/5#

Let #s=-5/3#, making both the #A# and #B# terms become #0#:

#1=C(-5/3)^2=C(25/9)#

#C=9/25#

Arbitrarily let #s=5# to solve for #A#, using #B=1/5# and #C=9/25#:

#1=A(5)(3*5+5)+1/5(3*5+5)+9/25(5)^2#

#1=A(100)+4+9#

#-12=100A#

#A=-3/25#

Thus:

#1/(s^2(3s+5))=-3/(25s)+1/(5s^2)+9/(25(3s+5))#

So:

#int1/(s^2(3s+5))ds=-3/25int1/sds+1/5ints^-2ds+9/25int1/(3s+5)ds#

Using typical integration rules (don't forget to substitute in the final integral):

#=-3/25ln(abss)-1/(5s)+3/25ln(abs(3s+5))+C#

Adam Adkins · Answer

undefined To integrate $ \frac{1}{{s^2(3s+5)}} $ using partial fractions, first express the fraction as a sum of simpler fractions:

$ \frac{1}{{s^2(3s+5)}} = \frac{A}{s} + \frac{B}{{s^2}} + \frac{C}{{3s+5}} $

To solve for A, B, and C, clear the denominators:

$ 1 = A(s)(3s+5) + Bs(3s+5) + Cs^2 $

Then, equate coefficients of like terms:

For $ s^0 $: $ A \times 0 = 1 \Rightarrow A = \frac{1}{5} $

For $ s^1 $: $ 3A + 3B = 0 \Rightarrow B = -\frac{1}{5} $

For $ s^2 $: $ 5A = C \Rightarrow C = \frac{3}{5} $

Now, integrate each term:

$ \int \frac{1}{{s^2(3s+5)}} ds = \int \frac{1}{5s} ds - \int \frac{1}{5s^2} ds + \int \frac{3}{5(3s+5)} ds $

$ = \frac{1}{5} \ln|s| + \frac{1}{5s} + \frac{1}{15} \ln|3s+5| + C $