You want to evaluate the following integral:

$I = \int_1^e \ln(x) \cdot x \cdot (\ln(x)^2 + 1) \, dx$

Step-by-step solution:

1. Expand the integrand:

   First, distribute the terms inside the integrand:

   $\ln(x) \cdot x \cdot (\ln(x)^2 + 1) = \ln(x) \cdot x \cdot \ln(x)^2 + \ln(x) \cdot x$

   So the integral becomes:

   $I = \int_1^e \left( \ln(x) \cdot x \cdot \ln(x)^2 + \ln(x) \cdot x \right) \, dx$

2. Split the integral:

   We can now separate the integral into two parts:

   $I = \int_1^e \ln(x) \cdot x \cdot \ln(x)^2 \, dx + \int_1^e \ln(x) \cdot x \, dx$

Solve each part:

First part: $\int_1^e \ln(x) \cdot x \cdot \ln(x)^2 \, dx$

To solve this, we use substitution. Let:

$u = \ln(x), \quad du = \frac{1}{x} \, dx$

Thus, the integral becomes:

$\int_1^e \ln(x) \cdot x \cdot \ln(x)^2 \, dx = \int_0^1 u^3 \, du$

Now integrate:

$\int_0^1 u^3 \, du = \frac{u^4}{4} \bigg|_0^1 = \frac{1}{4}$

Second part: $\int_1^e \ln(x) \cdot x \, dx$

Again, use the substitution $u = \ln(x)$ and $du = \frac{1}{x} \, dx$. The integral becomes:

$\int_1^e \ln(x) \cdot x \, dx = \int_0^1 u \, du$

Now integrate:

$\int_0^1 u \, du = \frac{u^2}{2} \bigg|_0^1 = \frac{1}{2}$

Final answer:

Now, we add both results together:

$I = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$

Thus, the value of the integral is:

$\boxed{\frac{3}{4}}$

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Would you like a further breakdown or explanation? Here are some follow-up questions to deepen understanding:

1. How does the substitution method work in detail, and when is it best to use it?
2. What other methods can we use to evaluate integrals like this without expanding the terms?
3. How does this method apply when the limits of integration are different (for example, $0$ to $1$ instead of $1$ to $e$)?
4. What is the relationship between logarithmic and exponential functions when dealing with integrals?
5. How do we solve integrals involving powers of logarithmic functions?

Tip: When integrating logarithmic functions, substituting $u = \ln(x)$ often simplifies the process, as the derivatives of logarithmic functions are neatly related to $\frac{1}{x}$.