Solve the ordinary differential equation. 
Solve the equation: $$x(2y+1)dx=y(x^2-3x+2)dy$$

To solve this, I first move $y(x^2-3x+2)dy$ to the LHS, then divide both sides by $(2 y + 1)(x^2 - 3 x + 2)$, the equation becomes:
$$\dfrac{x}{x^2-3x+2}dx-\dfrac{y}{2y+1}dy=0$$
After that, I integrate both sides: $$\int\dfrac{x}{x^2-3x+2}dx-\int\dfrac{y}{2y+1}dy=C$$
Which gives: $$2\ln|x-2|-\ln|x-1|+\dfrac{\ln|2y+1|-2y-1}{4}=C$$
Did I do this correctly?

There is a mistake when you divide both sides by $(2y+1)(x^2-3x+2)$.The correct form is
$$
\frac{x}{x^2-3x+2}\mathrm{d}x=\frac{y}{2y+1}\mathrm{d}y
$$
so integrating both sides gives
$$
2\ln |x-2|-\ln |x-1|=\frac{1}{4}\left( \left( 2y+1 \right) -\ln |2y+1|+C^\prime \right) 
$$
