How do you integrate #int frac{2x-4}{(x-4)(x+3)(x-6)} dx# using partial fractions?

Applying partial fraction decomposition:

#(2x-4)/((x-4)(x+3)(x-6)) = A/(x-4)+B/(x+3)+C/(x-6)#

#=> 2x-4 = A(x+3)(x-6) + B(x-4)(x-6) + C(x-4)(x+3)#

#=> 2x-4 = (A+B+C)x^2 + (-3A -10B -C)x +(-18A +24B -12C)#

Equating equivalent coefficients gives us the system

#{(A+B+C = 0), (-3A -10B - C = 2), (-18A +24B -12C = 4):}#

Solving, we get

#{(A = -2/7), (B=-10/63), (C = 4/9):}#

Thus, substituting back in,

#(2x-4)/((x-4)(x+3)(x-6)) = -(2/7)/(x-4)-(10/63)/(x+3)+(4/9)/(x-6)#

and so, integrating,

#int(2x-4)/((x-4)(x+3)(x-6))dx#

# = int(-(2/7)/(x-4)-(10/63)/(x+3)+(4/9)/(x-6))dx#

# = -(2/7)int1/(x-4)dx -(10/63)int1/(x+3)dx#
#+(4/9)int1/(x-6)dx#

#= -2/7ln|x-4| - 10/63ln|x+3| + 4/9ln|x-6|+C#

First, you need to write out the partial fractions. The denominator has already been factorized for you.

#frac{2x - 4}{(x - 4)(x + 3)(x - 6)} -= frac{A}{x - 4} + frac{B}{x + 3} + frac{C}{x - 6}#

Where #A#, #B# and #C# are constants to be determined. Note that the sign #-=# means that the equality holds true for all possible values of #x#. Get rid of the denominators by multiplying both sides with #(x - 4)(x + 3)(x - 6)#.

#2x - 4 -= A(x + 3)(x - 6) + B(x - 4)(x - 6) + C(x - 4)(x + 3)#

When #x = 4#

#2(4) - 4 = A(4 + 3)(4 - 6)#

#A = -2/7#

When #x = -3#

#2(-3) - 4 = B(-3 - 4)(-3 - 6)#

#B = -10/63#

When #x = 6#

#2(6) - 4 = C(6 - 4)(6 + 3)#

#C = 4/9#

Therefore,

#frac{2x - 4}{(x - 4)(x + 3)(x - 6)} -= -frac{2/7}{x - 4} - frac{10/63}{x + 3} + frac{4/9}{x - 6}#.

Now, we proceed with the integration.

#int frac{2x - 4}{(x - 4)(x + 3)(x - 6)} dx = int (-frac{2/7}{x - 4} - frac{10/63}{x + 3} + frac{4/9}{x - 6}) dx#

#= -2/7 int frac{1}{x - 4} dx - 10/63 int frac{1}{x + 3} dx + 4/9 int frac{1}{x - 6} dx#

#= -2/7 ln|x - 4| - 10/63 ln|x + 3| + 4/9 ln|x - 6| + C#,

where #C# is the constant of integration.