The given mathematical expression is:

$\int_1^2 \left(6 \cdot \frac{1}{x} + 2 \cdot e^x\right) \, dx$

To solve this definite integral, we will break it into two parts and solve each one separately.

Step 1: Separate the integral

$\int_1^2 \left(6 \cdot \frac{1}{x} + 2 \cdot e^x \right) \, dx = \int_1^2 6 \cdot \frac{1}{x} \, dx + \int_1^2 2 \cdot e^x \, dx$

Step 2: Solve the first integral

$\int 6 \cdot \frac{1}{x} \, dx = 6 \ln|x|$ Evaluating from $x = 1$ to $x = 2$:

$6 \ln|x| \Big|_1^2 = 6 \ln(2) - 6 \ln(1)$ Since $\ln(1) = 0$, this simplifies to: $6 \ln(2)$

Step 3: Solve the second integral

$\int 2 \cdot e^x \, dx = 2e^x$ Evaluating from $x = 1$ to $x = 2$:

$2e^x \Big|_1^2 = 2e^2 - 2e^1 = 2e^2 - 2e$

Step 4: Combine the results

Now, adding the two results together:

$6 \ln(2) + (2e^2 - 2e)$

Thus, the final solution to the integral is:

$6 \ln(2) + 2e^2 - 2e$

Would you like further details or clarifications?

Here are 5 related questions to expand your understanding:

1. How do we handle logarithmic functions in integrals?
2. What are the properties of the natural logarithm function?
3. How does the exponential function behave in definite integrals?
4. What are some applications of definite integrals in real-world problems?
5. Can you try solving a similar integral with different limits of integration?

Tip: Always check if the functions you're integrating have standard results. Many common integrals (like $e^x$ and $1/x$) have easily recognized antiderivatives.