$\int e^{2 x^2+\ln x} d x$ is equal to :

$\frac{e^{2x^2}}{4}+c$ $\frac{e^{2x^2}}{2}+c$ $\frac{e^{2x^2}}{4}+\frac{x^2}{2}$ $\frac{x e^{2x^2}}{4}+c$

$\frac{e^{2x^2}}{4}+c$

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.