Prove that: $$\int_{x}^{x+ \frac{ln x}{x}} e^{t^2} dt > \frac{e^{x^2} \ln x}{2 x}$$
Is this correct?
From The Mean Value Theorem for Definite Integrals:
$$\int_{x}^{x+ \frac{\ln x}{x}} e^{t^2} dt = \frac{\ln x}{x} e^{c^2}$$ for $c \in [x, x + \frac{\ln x}{x}]$.
Because $e^{t^2}$ is increasing we have:
$\frac{\ln x}{x} e^{c^2} \ge \frac{\ln x}{x} e^{x^2} > \frac{\ln x}{2 x} e^{x^2}$.
As another user already mentioned, you want $x>1$, since $\ln x$ would either be undefined or less than zero otherwise. While your proof works, we do not need the mean value theorem. Simply use that $e^{t^2}$ is an increasing function for $t>0$, then $$\int_x^{x+\frac{\ln(x)}{x}} e^{t^2} d t \geq \int_x^{x+\frac{\ln(x)}{x}} e^{x^2} d t =\frac{\ln(x)}{x}e^{x^2}>\frac{\ln(x)}{2x}e^{x^2}$$