Answer:
[tex]\frac{1}{6} (-\frac{3}{(2x+3)} -\ln x+\ln(2x+3))+C[/tex]
Step-by-step explanation:
We are given: [tex]\int\frac{-6}{x(4x+6)^2} dx[/tex]
[tex]x = \frac{3}{2}\tan t\\dx = \frac{3}{2}(\tan ^2 t +1) dt[/tex]
[tex]\int\frac{-6}{x(4x+6)^2} dx= \frac{1}{6} \int(-\frac{6}{(3+2x)^2} -\frac{1}{x} +\frac{2}{3+2x})dx=\\\\= \frac{1}{6} (-\frac{3}{(2x+3)} -\ln x+\ln(2x+3))+C[/tex]
