How do you integrate #1 / [ (1-2x)(1-x) ]# using partial fractions?

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Answer 1 · 2017-01-25T06:12:58

The answer is #=-ln(|1-2x|)+ln(|1-x|)+C#

Let's start the decomposition into partial fractions

#1/((1-2x)(1-x))=A/(1-2x)+B/(1-x)#

#=(A(1-x)+B(1-2x))/((1-2x)(1-x))#

The denominators are the same ; so, we compare the numerators

#1=A(1-x)+B(1-2x)#

Let #x=1#, #=>#, #1=-B#, #=>#, #B=-1#

Let #x=1/2#, #=>#, #1=1/2A#, #=>#, #A=2#

Therefore,

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

so,

#intdx/((1-2x)(1-x))=2intdx/(1-2x)-1intdx/(1-x)#

#=2ln(|1-2x|)/(-2)+ln(|1-x|)+C#