(4x-3)/x^3(x+1)=(Ax^2+Bx+C)/x^3+D/(x+1), so (Ax^2+Bx+C)(x+1)+Dx^3 must match 4x-3.
Ax^3+Ax^2+Bx^2+Bx+Cx+C+Dx^3
A+D=0 (x^3)
A+B=0 (x^2)
B+C=4 (x)
C=-3 (constant), so B=7, A=-7 and D=7
Partial fractions: (-7x^2+7x-3)/x^3+7/(x+1) or (7x-7x^2-3)/x^3+7/(x+1) or 7/(x+1)-(7x^2-7x+3)/x^3.