\begin{align} v^2 = A(v+2)(v+3) + B(v+1)(v+3) + C(V+2)(V+1)\end{align}

If \(v=-2\): \(4 = A(0) + B (-1) + C (0)\) : \(B = - 4\)

If \(v=-3\): \(9 = A(0) + B (0) + C (2)\) : \(C = 4.5\)

If \(v=-1\): \(1 = A(2) + B (0) + C (0)\) : \(A = \frac{1}{2}\)

\(\int{\frac{v^2 dv}{(v+1)(v+2)(v+3)}} = \int{\frac{Adv}{v+1}} + \int{\frac{Bdv}{v+2}} + \int{\frac{Cdv}{v+3}}\)

\(\int{\frac{v^2 dv}{(v+1)(v+2)(v+3)}} = \int{\frac{dv}{2(v+1)}} + \int{\frac{-4dv}{v+2}} + \int{\frac{9dv}{2(v+3)}}\)

\( \int{\frac{v^2dv}{(v+1)(v+2)(v+3)}} = \frac{1}{2} \ln(v+1) -4 \ln(v+2) + \frac{9}{2} \ln(v+3) +C\)