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? 

Partial fraction decomposition gives:
\begin{align*}
\dfrac{1}{y (3-y)} & = \dfrac{A}{y} + \dfrac{B}{3-y}\\
1&=A(3-y) + By
\end{align*}
Looking at the $y$ coefficients and the constant coefficients, we obtain the following two equations:
$$0 = A - B $$
$$1 = 3A$$
Which solve to give $A = B = \dfrac{1}{3}$.
Hence 
$$\dfrac{1}{y (3-y)} = \dfrac{1}{3y} + \dfrac{1}{3(3-y)}$$
or
$$\dfrac{1}{y (3-y)} = \dfrac{1}{3} \left(\dfrac{1}{y} + \dfrac{1}{3-y}\right)$$
