How do you integrate #int (13x) / (6x^2 + 5x - 6)# using partial fractions?

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Answer 1 · 2016-06-14T23:39:52

#2/3ln(abs(3x-2))+3/2ln(abs(2x+3))+C#

Note that

#6x^2+5x-6=6x^2+9x-4x-6#

#=3x(2x+3)-2(2x+3)#

#=(3x-2)(2x+3)#

The partial fraction decomposition can then be set up as:

#(13x)/((3x-2)(2x+3))=A/(3x-2)+B/(2x+3)#

Multiplying both sides by #(3x-2)(2x+3)#, we see that

#13x=A(2x+3)+B(3x-2)#

Letting #x=-3/2#:

#13(-3/2)=A(2(-3/2)+3)+B(3(-3/2)-2)#

#-39/2=A(-3+3)+B(-9/2-2)#

#-39/2=B(-13/2)#

#3=B#

Letting #x=2/3#:

#13(2/3)=A(2(2/3)+3)+B(3(2/3)-2)#

#26/3=A(4/3+3)+B(2-2)#

#26/3=A(13/3)#

#2=A#

Thus,

#(13x)/((3x-2)(2x+3))=2/(3x-2)+3/(2x+3)#

So,

#int(13x)/(6x^2+5x-6)dx=2int1/(3x-2)dx+3int1/(2x+3)dx#

#=2/3int3/(3x-2)dx+3/2int2/(2x+3)dx#

#=2/3ln(abs(3x-2))+3/2ln(abs(2x+3))+C#