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