Question 1193654
Certainly, let's tackle this integral using trigonometric substitution.

**1. Preparation**

* **Substitution:** 
    * Let ln(x) = 2sec(θ) 
    * Then, d(ln(x)) = 2sec(θ)tan(θ) dθ 
    * Which implies: (1/x) dx = 2sec(θ)tan(θ) dθ

**2. Substitute into the Integral**

* The integral becomes: 
   ∫ (ln³(x)) / (x)(√(ln²(x) - 4)) dx 
   = ∫ (2sec(θ))³ * 2sec(θ)tan(θ) / (√(4sec²(θ) - 4)) dθ 
   = ∫ 16sec⁴(θ)tan(θ) / (2√(sec²(θ) - 1)) dθ 
   = ∫ 8sec⁴(θ)tan(θ) / tan(θ) dθ 
   = ∫ 8sec⁴(θ) dθ

**3. Simplify the Integral**

* Use the identity: sec²(θ) = 1 + tan²(θ)
   ∫ 8sec⁴(θ) dθ = ∫ 8(1 + tan²(θ))sec²(θ) dθ 
   = ∫ 8sec²(θ) dθ + ∫ 8tan²(θ)sec²(θ) dθ

**4. Evaluate the Integrals**

* ∫ 8sec²(θ) dθ = 8tan(θ) + C₁ 
* ∫ 8tan²(θ)sec²(θ) dθ = 8/3 * tan³(θ) + C₂ 

**5. Combine the Results**

* ∫ 8sec⁴(θ) dθ = 8tan(θ) + 8/3 * tan³(θ) + C 

**6. Back-Substitute for x**

* Recall: ln(x) = 2sec(θ) 
   => sec(θ) = ln(x) / 2 
   => tan(θ) = √(sec²(θ) - 1) = √((ln²(x) / 4) - 1) = √(ln²(x) - 4) / 2

* Substitute these values back into the result:

   ∫ (ln³(x)) / (x)(√(ln²(x) - 4)) dx = 8 * (√(ln²(x) - 4) / 2) + 8/3 * (√(ln²(x) - 4) / 2)³ + C 
   = 4√(ln²(x) - 4) + (1/3) * √(ln²(x) - 4)³ + C

**Therefore, the solution to the integral is:**

∫ (ln³(x)) / (x)(√(ln²(x) - 4)) dx = 4√(ln²(x) - 4) + (1/3) * √(ln²(x) - 4)³ + C 

I hope this comprehensive solution is helpful!