Use the substitution $u=1+\sqrt{t}$ to show that the integral

$$\int \frac{12\sqrt{t}}{1+\sqrt{t}}dt$$

can be written in the form

$$\int \left(24u-48+\frac{24}{u}\right)du$$

The mass of a fungal colony, $m$ grams, grows at a rate modelled by the equation

$$\frac{dm}{dt}=\frac{12\sqrt{t}}{1+\sqrt{t}}$$

where $t$ is the number of days since the colony was first observed, for $1\le t\le 9$.

Determine the total increase in the mass of the colony from the end of day 1 to the end of day 9. Show each stage of your working and give your answer to one decimal place.