Help with writing the following as a partial fraction $\frac{4x+5}{x^3+1}$. I need help with writing the following as a partial fraction:
$$\frac{4x+5}{x^3+1}$$
My attempts so far are to factor $x^3$ into $(x+1)$ and $(x^2-x+1)$
This gives me: $A(x^2-x+1)+B(x+1)$. 
But I have problems with solving the equation system that this gives:
$A = 0$ (since there are no $x^2$ terms in $4x+5$)
$-A+B =4$ (since there are $4$ $x$ terms in $4x+5$)
$A+B = 5$ (since the constant is $5$ in $4x+5$)
this gives me $A=0.5$ and $B=4.5$ and $\frac{1/2}{x+1}, \frac{9/2}{x^2-x+1}$
This is appearantly wrong. Where is my reasoning faulty?
Thank you!

You need to use one less exponent per factor in the numerator after your factorization.
This leads to:
$$\frac{Ax+B}{x^2-x+1} + \frac{C}{x+1} = \frac{4x+5}{x^3+1}$$
This gives us:
$$Ax^2 + Ax + Bx + B + Cx^2 - Cx + C = 4x + 5$$
This leads to:
$A + C = 0$
$A + B - C = 4$
$B + C = 5$
yielding:
$$A = -\frac{1}{3}, B = \frac{14}{3}, C = \frac{1}{3}$$
Writing the expansion out yields:
$$\frac{4x+5}{x^3+1} = \frac{14 - x}{3(x^2-x+1)} + \frac{1}{3(x+1)}$$
