Step 1: Plan.
Instead of guessing a function, we will use integration by parts and see what remains. Split the integral as $\int e^{x}\log x\,dx+\int \frac{e^{x}}{x^{2}}dx$.
Step 2: Parts on the first piece.
Take $u=\log x$ and $dv=e^{x}dx$. Then $du=\frac{dx}{x}$ and $v=e^{x}$.
\[ \int e^{x}\log x\,dx = e^{x}\log x-\int \frac{e^{x}}{x}dx \]
Step 3: Parts on the second piece.
Take $u=e^{x}$ and $dv=\frac{dx}{x^{2}}$. Then $du=e^{x}dx$ and $v=-\frac{1}{x}$.
\[ \int \frac{e^{x}}{x^{2}}dx = -\frac{e^{x}}{x}+\int \frac{e^{x}}{x}dx \]
Step 4: Add the two results.
The unknown integral $\int \frac{e^{x}}{x}dx$ appears with opposite signs, so it cancels.
\[ \int e^{x}\left(\log x+\frac{1}{x^{2}}\right)dx = e^{x}\log x-\frac{e^{x}}{x}+c \]
Step 5: Match with the options.
We can write the result as $e^{x}\left(\log x-\frac{1}{x}\right)+c$. Option 3 has exactly this form. Options 1, 2 and 4 have a different sign or miss a term, so they do not match.
Final Answer:
Option 3 is correct. \[ \boxed{e^{x}\left(\log x-\frac{1}{x}\right)+c} \]