How do I compute the partial fraction decomposition of $\frac{6}{(x^4)(x+1)}$?

I let $$\frac{6}{(x^4)(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x^4} + \frac{F}{x+1}$$

When I let $x=-1$, $F=6$ and when $x=0$, $D=6$. How can I find the constants $A, B, and C?$.

Thanks.

How do I compute the partial fraction decomposition of $\frac{6}{x^4(x+1)}$?

I let $$\frac{6}{x^4(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x^4} + \frac{F}{x+1}$$

When I let $x=-1$, $F=6$ and when $x=0$, $D=6$. How can I find the constants $A, B,$ and $C?$.

Thanks.