Divide the following complex numbers
\(\frac{(3+i)}{(2-3i)}=\frac{(3+i)(2+3i)}{(2-3i)(2+3i)}=\frac{6+9i+2i+3i^2}{2^2-(3i)^2}=\frac{6+11i-3}{4-9i^2}=\frac{3+11i}{4+9}=\frac{3+11i}{13}=\\ =\boxed{\frac{3}{13}+ \frac{11}{13}i} \Rightarrow \hbox{the answer is D}\)
