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,

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

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

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

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