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 $int1/(s(s-1)^2)ds=int1/sds-int1/(s-1)ds+int1/(s-1)^2ds$

$lns-ln(s-1)-1/(s-1)$

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