Integrating a function with three products

Question

I'm trying to integrate this function:
$$\frac {dy}{dt}=k\left(n_1-{\frac{y}{2}}\right)^2\left(n_2-{\frac{y}{2}}\right)^2\left(n_3-{\frac{3y}{4}}\right)^3$$
I know it's $$\frac {dy}{dt}=k\int\left(3320-{\frac{y}{2}}\right)^2\left(3265-{\frac{y}{2}}\right)^2\left(3581-{\frac{3y}{4}}\right)^3$$
but I'm unsure where to go on from there. Would solving the first product be $2\frac{(n_2-\frac{x}{2})^3}{3}$?

Answer 1

To make life easier, start changing notation to get $$\frac{dy}{dt}=\frac {27 k}{1024} (a-y)^2(b-y)^2(c-y)^3$$ where $a=2n_1$, $b=2n_2$ and $c=\frac 43n_3$.

As Michael Burr commented, use partial fraction decomposition and get $$\frac 1 {(a-y)^2(b-y)^2(c-y)^3}=\frac{5 a-3 b-2 c}{(a-b)^3 (a-c)^4 (y-a)}+\frac{3 a-5 b+2 c}{(a-b)^3 (b-c)^4 (y-b)}+\frac{-3 a^2-4 a b+10 a c-3 b^2+10 b c-10 c^2}{(a-c)^4 (c-b)^4 (y-c)}+\frac{2 (a+b-2 c)}{(a-c)^3 (c-b)^3 (y-c)^2}-\frac{1}{(a-b)^2 (a-c)^3 (y-a)^2}-\frac{1}{(a-b)^2 (b-c)^3 (y-b)^2}-\frac{1}{(a-c)^2 (c-b)^2 (y-c)^3}$$ and you face quite simple integrals