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

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Answer 1 · 2016-12-20T08:23:14

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

Let's rewrite the expression

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

The decomposition into partial fractions is

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

#=(A(x-1)^2(x+1)^2+Bx(x+1)^2+Cx(x-1)(x+1)^2+Dx(x-1)^2+Ex(x-1)^2(x+1))/(x(x-1)^2(x+1)^2)#

Therefore,

#1=A(x-1)^2(x+1)^2+B(x)(x+1)^2+Cx(x-1)(x+1)^2+Dx(x-1)^2+Ex(x-1)^2(x+1)#

Let #x=0#, #=>#, 1=A#

Let #x=1#, #=>#, #1=4B#, #=>#, #B=1/4#

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

Coefficients of #x^4#, #=>#,#0=A+C+E#, #=>#, #C+E=-1#

Coeficients of #x^3#, #=>#, #0=B+C+D-E#, #=>#, #C-E=0#

#C=E=-1/2#

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