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

1 Answer
Nov 30, 2016

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

Explanation:

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)