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

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Answer 1 · 2018-03-10T21:08:48

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

#(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#