How do you integrate #int x/((x-6)(x-3) dx# using partial fractions?

1 Answer
Nov 24, 2015

#intx/((x-6)(x-3))dx = 2ln|x-6| - ln|x-3| + C#

Explanation:

To find the partial fraction decomposition, we begin with

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

Multiplying by #(x-6)(x-3)# gives us

#x = A(x-3)+B(x-6) = (A+B)x + (-3A - 6B)#

Matching respective coefficients on each side gives us the system of equations

#{(A + B = 1), (-3A -6B = 0):}#

Now all we need to do is solve for #A# and #B#. For example, using elimination,

#(-3A - 6B) + 3(A+B) = 0 + 3(1)#
#=> -3B = 3#
#=> B = -1#

Then, substituting

#A - 1 = 1#
#=> A = 2#

So the decomposition is

#x/((x-6)(x-3)) = 2/(x-6) - 1/(x-3)#

Now we can do the integral relatively easily.

#intx/((x-6)(x-3))dx = int(2/(x-6) - 1/(x-3))dx#

#int(2/(x-6) - 1/(x-3))dx = 2int(1/(x-6))dx - int(1/(x-3))dx#

A couple of simple #u# substitutions, together with #int(1/x)dx = ln|x|+C#
gives us

#2int(1/(x-6))dx = 2ln|x-6| + C#
and
#-int(1/(x-3))dx = -ln|x-3|+C#

So putting it all together, we get

#intx/((x-6)(x-3))dx = 2ln|x-6| - ln|x-3| + C#