How do you take the integral of $\int \frac{\mathrm{d}y}{y(3-y)}$? How do you evaluate the following? $$\int \frac{\mathrm{d}y}{y(3-y)}$$ 
I have looked at the solution and I don't understand how they are taking the integral of this? They go from: 
$$\int \frac{\mathrm{d}y}{y(3-y)}$$
to
$$\int \frac13 \left(\frac1y + \frac{1}{3-y}\right)\, \mathrm{d}y$$
and how they get to this second step confuses me. Where did they get this extra $\frac13$ from? 

$\int\frac{1}{y(3-y)}dy$ can written into a more integrable form using partial fractions as given below:
$$\frac{1}{y(3-y)} = \frac{A}{y} + \frac{B}{3-y}$$
$$1 \equiv A(3-y) + By$$
$$A = B = \frac{1}{3}$$
$$\int\frac{1}{y(3-y)} \, \mathrm{d}y = \int\left(\frac{1}{3y} + \frac{1}{3(3-y)}\right)dy = \int \frac{1}{3}\left(\frac{1}{y} + \frac{1}{3-y}\right) \, \mathrm{d}y$$
You can then integrate the above wrt $y$
