Answer:
Option C
Explanation:
$\frac{dy}{dx}+\frac{y}{x\log_{}{}x} =\frac{\sin2x}{\log_{}{}x}$
I.F. = $e^{\int_{}^{}\frac{dx}{x\log_{}{}x} }$
.'. I.F. = $e^{\int_{}^{}\frac{1}{t}dt}$ = $e^{\log_{}{}t}$ = t = log |x|
solution is given by
y(I.F.) = $\int_{}^{}Q(I.F.)dx + C$
y log |x| = $\int_{}^{}\frac{sin 2x}{log |x|}(log|x|)dx +C = -\frac{Cos 2x}{2}+C $