How do you integrate int (4x^2 - 7x - 12) / ((x) (x+2) (x-3)) using partial fractions?

1 Answer
Mar 10, 2018

2ln absx +9/5lnabs(x+2)+1/5lnabs(x-3)+ C

Explanation:

(4x^2-7x-12)/((x)(x+2)(x-3))=A/x+B/(x+2)+C/(x-3)=

=(A(x+2)(x-3)+B(x)(x-3)+C(x)(x+2))/((x)(x+2)(x-3))
then:

A(x^2-x-6)+B(x^2-3x)+C(x^2+2x)=4x^2-7x-12

cons. x=0 then A(-6)+B(0)+C(0)=-12 rArr A=2

x^2-x-6=0rArrx_1=3and x_2=-2

cons. x=-2 then A(0)+B(10)+C(0)=18 rArr B=9/5

cons. x=3 then A(0)+B(0)+C(15)=3 rArr C=1/5

finally get to:

(4x^2-7x-12)/((x)(x+2)(x-3))=2/x+9/(5(x+2))+1/(5(x-3))

then:

int(4x^2-7x-12)/((x)(x+2)(x-3))dx=int(2/x+9/(5(x+2))+1/(5(x-3)))dx=int2/xdx+int9/(5(x+2))dx+int1/(5(x-3))dx=

2ln absx +9/5lnabs(x+2)+1/5lnabs(x-3)+ C