To solve the integral
\[
I = \int \frac{(\ln x)^5}{\sqrt{x^2 + x^2 (\ln x)^3}} \, dx,
\]
we will follow a series of steps to simplify and evaluate the integral.
### Step 1: Simplify the Integral
First, we can factor out \(x^2\) from the square root in the denominator:
\[
\sqrt{x^2 + x^2 (\ln x)^3} = \sqrt{x^2(1 + (\ln x)^3)} = x\sqrt{1 + (\ln x)^3}.
\]
Thus, we can rewrite the integral as:
\[
I = \int \frac{(\ln x)^5}{x\sqrt{1 + (\ln x)^3}} \, dx.
\]
### Step 2: Substitution
Now, we will use the substitution \(t = \ln x\). Then, we have:
\[
dx = e^t \, dt \quad \text{and} \quad x = e^t.
\]
Substituting these into the integral gives:
\[
I = \int \frac{t^5}{e^t \sqrt{1 + t^3}} e^t \, dt = \int \frac{t^5}{\sqrt{1 + t^3}} \, dt.
\]
### Step 3: Further Substitution
Next, we can make another substitution. Let \(u = t^3\), then \(du = 3t^2 \, dt\), or \(dt = \frac{du}{3t^2}\). We also have \(t = u^{1/3}\), so:
\[
t^5 = (u^{1/3})^5 = u^{5/3} \quad \text{and} \quad \sqrt{1 + t^3} = \sqrt{1 + u}.
\]
Substituting these into the integral gives:
\[
I = \int \frac{u^{5/3}}{\sqrt{1 + u}} \cdot \frac{du}{3u^{2/3}} = \frac{1}{3} \int \frac{u}{\sqrt{1 + u}} \, du.
\]
### Step 4: Integrate
Now we can simplify the integral:
\[
I = \frac{1}{3} \int u^{1/2} (1 + u)^{-1/2} \, du.
\]
This integral can be solved using the substitution \(v = 1 + u\), leading to:
\[
du = dv \quad \text{and} \quad u = v - 1.
\]
Thus, we have:
\[
I = \frac{1}{3} \int (v - 1)v^{-1/2} \, dv = \frac{1}{3} \left( \int v^{1/2} \, dv - \int v^{-1/2} \, dv \right).
\]
Calculating these integrals gives:
\[
I = \frac{1}{3} \left( \frac{2}{3} v^{3/2} - 2v^{1/2} \right) + C.
\]
### Step 5: Back Substitution
Now substituting back \(v = 1 + t^3\) and \(t = \ln x\):
\[
I = \frac{1}{3} \left( \frac{2}{3} (1 + (\ln x)^3)^{3/2} - 2(1 + (\ln x)^3)^{1/2} \right) + C.
\]
### Step 6: Compare with Given Form
The problem states that:
\[
I = k \sqrt{(\ln x)^3 + 1} \left( (\ln x)^3 - 2 \right) + C.
\]
By comparing coefficients, we find:
\[
k = \frac{2}{9}.
\]
### Step 7: Calculate \(9k\)
Finally, we calculate:
\[
9k = 9 \cdot \frac{2}{9} = 2.
\]
Thus, the answer is:
\[
\boxed{2}.
\]
