How do you express #(x^4)/((x²+1)(x²-1))# in partial fractions?

Question

5465 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-30T18:19:21

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

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

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

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

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

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

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

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

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

No coefficients, #1=-B-C+D#, #=>#, #B=1/2-1=-1/2#

Coefficients of #x^3#, #=>#, #0=A+C+D#, #=>#. #A=0#

Therefore,

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