integral of 1/(w^4-w^3)
1/(w^4 – w^3) = 1/{w^3(w – 1)} = (Aw^2 + Bw + C)/w^3 + D/(w – 1)
= (Aw^3 + Bw^2 + Cw – Aw^2 – Bw – C +Dw^3)/{w^3(w – 1)}
= {(A + D)w^3 + (B – A)w^2 + (C – B)w – C}/(w^4 – w^3)
Equating coefficient values,
A + D = 0
B – A = 0
C – B = 0
-C = 1
All of which give: A = B = C = -1, D = 1
Therefore, 1/(w^4 – w^3) = (-w^2 – w – 1)/w^3 + 1/(w – 1)
1/(w^4 – w^3) = -1/w – 1/w^2 – 1/w^3 + 1/(w – 1)
So,
Int[1/(w^4 – w^3)]dw = int[-1/w – 1/w^2 – 1/w^3 + 1/(w – 1)]dw
= int[-1/w]dw – int[1/w^2]dw – int[1/w^3]dw + int[1/(w – 1)]dw
= -ln(w) + 1/w + 1/(2w^2) + ln(w – 1)