∫ e^2 x^2+ln x d x is equal to :

Let $I=\int e^{2 x^2} d x$

Let $x^2=t \Rightarrow 2x~dx=dt$

$\Rightarrow I=\frac{1}{2} \int e^{2 t} d t=\frac{e^{2 t}}{4}+c=\frac{e^{2 x^2}}{4}+c$

Hence (1) is the correct answer.