How do you integrate #int (2x-2)/((x-4)(x-1)(x-6)) # using partial fractions?

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Answer 1 · 2016-11-15T08:19:36

The answer is #=-ln(x-4)+ln(x-6) +C#

Let's start the decomposition into partial fractions

#(2x-2)/((x-4)(x-1)(x-6))=(2cancel(x-1))/((x-4)cancel(x-1)(x-6))#

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

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

So, #2=(A(x-6)+B(x-4))#

Let #x=4# #=>##2=-2A# #=>##A=-1#

Let #x=6# #=># #2=2B# #=># #B=1#

#2/((x-4)(x-6))=-1/(x-4)+1/(x-6)#

#int(2dx)/((x-4)(x-6))=int(-1dx)/(x-4)+int(1dx)/(x-6)#

#=-ln(x-4)+ln(x-6) +C#