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#