How do you integrate $int (2s)/[(s+4)(s-1)]$ using partial fractions?

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Answer 1 · 2016-05-30T16:14:47

$8/5ln(abs(s+4))+2/5ln(abs(s-1))+C$

Split up $(2s)/((s+4)(s-1))$ into its partial fraction decomposition:

$(2s)/((s+4)(s-1))=A/(s+4)+B/(s-1)$

$2s=A(s-1)+B(s+4)$

Letting $s=1$ :

$2(1)=A(1-1)+B(1+4)$

$2=5B$

$B=2/5$

Letting $s=-4$ :

$2(-4)=A(-4-1)+B(-4+4)$

$-8=-5A$

$A=8/5$

Thus,

$(2s)/((s+4)(s-1))=8/5(1/(s+4))+2/5(1/(s-1))$

Splitting up the integral through addition:

$int(2s)/((s+4)(s-1))ds=8/5int1/(s+4)ds+2/5int1/(s-1)ds$

Both of these are simply integrated through the natural logarithm:

$=8/5ln(abs(s+4))+2/5ln(abs(s-1))+C$

How do you integrate int (2s)/[(s+4)(s-1)]int (2s)/[(s+4)(s-1)] using partial fractions?