How do you write the partial fraction decomposition of the rational expression #1/(x^4-1)#?

1 Answer
George C.
Jan 2, 2016

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

#= 1/(4(x-1))-1/(4(x+1))+i/(4(x-i))-i/(4(x+i))#

Explanation:

Sticking with Real coefficients for now:

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

So we want to solve:

#1/(x^4-1) = A/(x-1)+B/(x+1)+(Cx+D)/(x^2+1)#

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

#=((A(x+1)+B(x-1))(x^2+1)+(Cx+D)(x^2-1))/(x^4-1)#

#=((A+B+C)x^3+(A-B+D)x^2+(A+B-C)x+(A-B-D))/(x^4-1)#

So:

(i) #A+B+C=0#

(ii) #A-B+D=0#

(iii) #A+B-C=0#

(iv) #A-B-D=1#

If we subtract (iii) from (i) we find #2C = 0#, so #C=0# and:

(v) #A+B=0#

If we subtract (iv) from (ii) we find #2D = -1#, so #D = -1/2# and:

(vi) #A-B=1/2#

Adding (v) and (vi) we find #2A=1/2#, so #A=1/4#, hence #B=-1/4#

So:

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

If we allow Complex coefficients then we find:

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

So:

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

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