\int\frac{2x-3}{(2x+3)^4}dx=
2x+3=t
2x=t-3
2dx=dt
dx=\frac{dt}{2}
=\frac{1}{2}\int\frac{t-6}{t^4}dt=\frac{1}{2}\int\frac{1}{t^3}dt-3\int\frac{1}{t^4}dt=\frac{1}{2*(-2)t^2}-3\frac{1}{(-3)t^3}+C=
=-\frac{1}{4t^2}+\frac{1}{t^3}+C=\frac{-t+4}{4t^4}+C=
=\frac{-2x-3+4}{4(2x+3)^3}+C=\frac{-2x+1}{4(2x+3)^3}+C