How do you integrate #int (2s)/[(s+4)(s-1)]# using partial fractions?
1 Answer
May 30, 2016
Explanation:
Split up
#(2s)/((s+4)(s-1))=A/(s+4)+B/(s-1)#
#2s=A(s-1)+B(s+4)#
Letting
#2(1)=A(1-1)+B(1+4)#
#2=5B#
#B=2/5#
Letting
#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#