Given Newton's 2nd law equation, I'm supposed to find $v$. My second law equation is: $m\dot{v}=-bv-vc^2$
By separation of variables, I arrive at $$\frac{dv}{bv+cv^2}=\frac{-1}{m}dt$$
$$\int_{v_0}^{v}\frac{dv}{bv+cv^2}=\frac{-1}{m}\int_0^tdt$$
To solve my integral on the left, my professor told us that partial fraction decomp will be the easiest way of solving. $$\frac{1}{v(b+cv)}=\frac{A}{v}+\frac{B}{b+cv}$$
$$1=A(b+cv)+B(v)$$ I believe I can find one of the numerators by setting $v=0$, that is: $$1=A(b+c\cdot0)+B(0)$$
$$1=Ab$$
$$\therefore A=\frac{1}{b}$$
I can also make $v=\frac{1}{c}$ (assuming this is allowed), in which case:
$$1=A(b+c\cdot\frac{1}{c})+B(\frac{1}{c})$$
$$1=\frac{1}{b}(b+1)+\frac{B}{c}$$
$$1=1+\frac{1}{b}+\frac{B}{c}$$
$$\therefore B=-\frac{c}{b}$$
My integral can now be written as:
$$\int_{v_0}^{v}\frac{dv}{bv+cv^2}=\frac{1}{b}\int_{v_0}^v\frac{1}{v}-\frac{c}{b+cv}dv$$
In which case my solution is
$$\frac{1}{b}[\ln{v}-\ln{(b+cv)}]|_{v_0}^v$$
I want to confirm my work, as I didn't start making progress on the problem until my professor was no longer available, and my other classmates are also confused.