Question:

Put $4x = 3t \Rightarrow 4dx = 3dt$<br>$\therefore \int \frac{dx}{9-16x^2} = \int \frac{3dt}{4[9-9t^2]} = \frac{-1}{12} \int \frac{dt}{t^2-1} = \frac{1}{24} \ln \left| \frac{1+t}{1-t} \right| + C = \frac{1}{24} \ln \left| \frac{3+4x}{3-4x} \right| + C$