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

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Answer 1 · 2016-09-08T09:09:49

#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)#