16x=16(x-3)+48, so f(x)=16x/(x-3)=16+48/(x-3), and ∫f(x)dx=∫16dx/(x-3)=∫16dx+48∫dx/(x-3). Integral is 16x+48ln|x-3|+C or 16x+48ln|a(x-3)| where a and C are constants.
218-x=e(18-x)ln(2)=(e18ln(2))(e-xln(2)).
Let a=⅕(e18ln(2)) and b=-ln(2) or ln(0.5); so a and b are constants.
∫aebxdx=a∫ebxdx=(a/b)ebx+C, where C is a constant.
Therefore ∫f(x)dx=⅕(e18ln(2)/ln(0.5))exln(0.5)+C.
We can write this as: ⅕(218/ln(0.5))(0.5x)+C or -218-x/(5ln(2))+C.
5ln(2)=ln(25)=ln(32), so -218-x/(ln(32))+C is another way of expressing the solution.
I hope this is helpful and shows you how to simplify what looks terrifying! (I hope I got it right!)