Question

Decompose into partial fraction:
(x^3+1)/(x^3+x)

Answer 1

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

Answer 2

- this is possibile descompusing again ?

Answer 3

\[\frac{x^3+1}{x^3+x}=1+\frac{1-x}{x^3+x}\]
\[\frac{1-x}{x^3+x}=\frac{1-x}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}\]
\[1-x=A(x^2+1)+(Bx+C)x\]
\[1-x=x^2(A+B)+Cx+A\]
Therefore, A=1, B=-1, and C=-1.
Therefore, \[\frac{x^3+1}{x^3+x}=1+\frac{1}{x}-\frac{x+1}{x^2+1}\]

Answer 4

Ok, I'm assuming A+B = 0?

Answer 5

Yes, because what I did was equate coefficients. The coefficients of x^2 are 0 and A+B. So I assumed they were equal. :D

Answer 6

Thanks so much.

Answer 7

Glad to be of help. :D