How to solve integral of f(x)= (16x)/(x-3) and f(x)=(1/5)*2^(18-x)

Question

How do you solve those integrals?

f(x)=(16x)/(x-3)

f(x)=(1/5)*2^18-x

Answer 1

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.

2^18-x=e^(18-x)ln(2)=(e^18ln(2))(e^-xln(2)).

Let a=⅕(e^18ln(2)) and b=-ln(2) or ln(0.5); so a and b are constants.

∫ae^bxdx=a∫e^bxdx=(a/b)e^bx+C, where C is a constant.

Therefore ∫f(x)dx=⅕(e^18ln(2)/ln(0.5))e^xln(0.5)+C.

We can write this as: ⅕(2¹⁸/ln(0.5))(0.5^x)+C or -2^18-x/(5ln(2))+C.

5ln(2)=ln(2⁵)=ln(32), so -2^18-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!)