Evaluate.
\[
\int_{1}^{e^{3}}\left(3 x+\frac{1}{x}\right) d x
\]
$\int_{1}^{e^{3}}\left(3 x+\frac{1}{x}\right) d x=\square$ (Type an exact answer in terms of e.)

Step 1 ：Split the integral into two separate integrals: \(\int_{1}^{e^{3}} 3x dx\) and \(\int_{1}^{e^{3}} \frac{1}{x} dx\).

Step 2 ：Solve the first integral using the power rule to get the antiderivative \(\frac{3x^{2}}{2}\).

Step 3 ：Solve the second integral using the logarithmic rule to get the antiderivative \(\ln(x)\).

Step 4 ：Substitute the upper and lower limits of the integral into the antiderivatives to get \(\frac{3e^{6}}{2} - \frac{3}{2}\) and \(3\ln(e^{3}) - \ln(1)\).

Step 5 ：Simplify the results to get \(\frac{3e^{6}}{2} - \frac{3}{2}\) and \(3\cdot3 - 0\).

Step 6 ：Add the results of the two integrals to get the final answer: \(\boxed{\frac{3}{2} + \frac{3e^{6}}{2}}\).