How do you express #x/((x^2-1)(x-1))# in partial fractions?

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Answer 1 · 2017-02-16T08:07:03

The answer is #=(-1/4)/(x+1)+(1/2)/(x-1)^2+(1/4)/(x-1)#

Let's factorise the denominator

#x^2-1=(x+1)(x-1)#

Therefore,

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

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

Let's perform the decomposition into partial fractions

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

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

The denominators are the same, we compare the numerators

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

Let #x=-1#, #=>#, #-1=4A#, #=>#, #A=-1/4#

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

Coefficients of #x^2#

#0=A+C#, #=>#, #C=-A=1/4#

Therefore,

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