Solution:

$\frac{x+3}{(x-1)()x+1)(x+5)}$ = $\frac{A}{x-1}$ + $\frac{B}{x+1}$ + $\frac{C}{x+5}$

therefore, x+3 = A(x+1)(x+5) +B (x-1)(x+5) +C(x-1)(x+1)

put x = 1

4=A(2)(6)

4=12A

hence, A = $\frac{1}{3}$

put x = -1

-1 + 3 = B(-2)(4)

2 = -8B

hence, B = $-\frac{1}{4}$

put x = -5

-5 + 3 = C(-6)(-4)

-2 = 24C

hence, C = $\frac{-1}{12}$

$\frac{x+3}{(x-1)()x+1)(x+5)}$ = $\frac{\frac{1}{3}}{x-1}$ + $\frac{-\frac{1}{4}}{x+1}$ + $\frac{-\frac{1}{12}}{x+5}$