How do you use partial fraction decomposition to decompose the fraction to integrate #(13x) / (6x^2 + 5x - 6)#?

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Answer 1 · 2015-08-15T22:13:37

Using a fraction whose denominator is quadratic, you need to check whether or not it is factorable.

#6x^2 + 5x - 6 = (2x+3)(3x-2)#

Since it is, we only need to write out the following:

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

To find what #A# and #B# are, you could start by cross-multiplying. For now, ignore the integral sign and #dx#.

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

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

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

Thus, we have the equations:

#-2A + 3B = 0 => B = 2/3 A#

#3A + 2B = 13 => 3A + 4/3 A = 13#
#13/3 A = 13#
#:. color(green)(A = 3) => color(green)(B = 2)#

At this point we are pretty much done.

#= int 3/(2x + 3) + 2/(3x - 2)dx#

#= color(blue)(3/2ln|2x + 3| + 2/3ln|3x - 2| + C)#