for the second one:

the integral of 1/(9-5x)dx

first, let u = 9-5x

then, du = -5

so, -1/5du = dx

substitute

(-1/5) times integral of 1/u du

which is easy to integrate

(-1/5) times ln(u) + c

substitute back for u

so the answer is:

(-1/5)(ln(5x-9)) + C

edit: forgot the + C