How do you take the integral of $\int \frac{\mathrm{d}y}{y(3-y)}$?

Question

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?

Answer 1

$\int\frac{1}{y(3-y)}dy$ 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 $y$

Answer 2

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 $$\dfrac{1}{y (3-y)} = \dfrac{1}{3} \left(\dfrac{1}{y} + \dfrac{1}{3-y}\right)$$