How do you integrate #int (5x^2+3x-2)/(x^3+2x^2)# using partial fractions?

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Answer 1 · 2016-07-18T06:37:38

#1/x+ln|x^2(x+2)^3|+C_1#

The #Dr.=x^3+2x^2=x^2(x+2)#

So, we are having linear repeated factors. In such case, we take,

#(5x^2+3x-2)/(x^3+2x^2)=A/x+B/x^2+C/(x+2)....(1), where, A,B,C in RR#

#=(Ax(x+2)+B(x+2)+Cx^2)/(x^2(x+2))#

#rArr 5x^2+3x-2=Ax(x+2)+B(x+2)+Cx^2..............(2)#

Eventhough #(2)# is true #AAx in RR#, we select,

#x=0 rArr -2=2B rArr B=-1#

#x=-2 rArr20-6-2=12=4C rArr C=3#

#x=1 rArr 5+3-2=6=3A+3B+C=3A-3+3=3A#

#rArrA=2#

#:. I=int(5x^2+3x-2)/(x^3+2x^2)dx#

#=int2/xdx-int1/x^2dx+int3/(x+2)dx#

#=2lnx+1/x+3ln|x+2|#

#=1/x+ln|x^2(x+2)^3|+C_1#

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