The problem is asking to evaluate the improper integral:

$\int_e^{\infty} \frac{dx}{x \ln x \left(\ln^3 x + 1\right)}.$

Step-by-step solution:

1. Substitution:

Let: $u = \ln x, \quad \text{so } du = \frac{1}{x} dx.$

The limits of integration change as follows:

• When $x = e$, $u = \ln(e) = 1$.
• When $x \to \infty$, $u \to \infty$.

This transforms the integral into: $\int_{1}^{\infty} \frac{du}{u \left(u^3 + 1\right)}.$

2. Partial fraction decomposition:

We decompose the denominator $u \left(u^3 + 1\right)$. Notice that: $u^3 + 1 = (u + 1)(u^2 - u + 1).$

Thus: $\frac{1}{u \left(u^3 + 1\right)} = \frac{A}{u} + \frac{B}{u + 1} + \frac{Cu + D}{u^2 - u + 1},$ where $A$, $B$, $C$, and $D$ are constants to determine.

3. Find coefficients:

Expanding and equating terms gives: $\frac{1}{u \left(u + 1\right)\left(u^2 - u + 1\right)} = \frac{A}{u} + \frac{B}{u+1} + \frac{Cu + D}{u^2 - u + 1}.$

Multiply through by the denominator $u \left(u + 1\right)\left(u^2 - u + 1\right)$: $1 = A(u + 1)(u^2 - u + 1) + B u (u^2 - u + 1) + (Cu + D)u(u + 1).$

Expanding this equation and solving for $A$, $B$, $C$, and $D$, we find:

• $A = 1$,
• $B = -1$,
• $C = 1$,
• $D = 0$.

Thus, the partial fraction decomposition is: $\frac{1}{u \left(u^3 + 1\right)} = \frac{1}{u} - \frac{1}{u + 1} + \frac{u}{u^2 - u + 1}.$

4. Rewrite the integral:

Using the decomposition, the integral becomes: $\int_{1}^{\infty} \frac{1}{u} \, du - \int_{1}^{\infty} \frac{1}{u+1} \, du + \int_{1}^{\infty} \frac{u}{u^2 - u + 1} \, du.$

5. Evaluate each term:

• First term: $\int_{1}^{\infty} \frac{1}{u} \, du = \ln u \Big|_{1}^{\infty}.$ As $u \to \infty$, $\ln u \to \infty$. Hence, this term diverges.

6. Conclusion:

Since the first term diverges, the given integral is divergent.

Final Answer:

$\boxed{\text{The integral diverges.}}$

Would you like further details or clarification on any part? Here are 5 related questions for further exploration:

1. What are the typical criteria for determining whether improper integrals converge or diverge?
2. How can partial fraction decomposition be used effectively in other integrals?
3. What are some methods to evaluate convergent improper integrals?
4. Could a different substitution simplify the integral further?
5. How does the growth of the logarithm function contribute to divergence?

Tip: Always check for divergence of improper integrals by analyzing the growth of terms in the integrand at the limits of integration!