Answer:
Factorise and split the denominator:
1/(x^3 + 8x) = 1/(x(x^2+8)) = (1/x) * (1/(x^2+8))
now say
(a/x) + ((bx+c)/(x^2+8)) = (x+16)/(x^3+8x)
so
(x^2+8)a + (bx+c)x = x + 16
a must be 2 to get the + 16
2x^2 + 16 + bx^2 + cx = x + 16
b must be -2 to get rid of the 2x^2
c must be 1 to get the x
[tex]\frac{2}{x} + \frac{1-2x}{x^2+8}[/tex]
