$$ \int _ { 3 } ^ { 6 } \frac { \ln x } { ( x - 3 ) ^ { 4 } } d x = $$

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To evaluate the integral:

$$
\int_{3}^{6} \frac{\ln x}{(x - 3)^4} \, dx
$$

let's analyze its behavior and determine whether it converges or diverges.

### Step 1: Substitute Variables

Let $$ t = x - 3 $$. Then $$ x = t + 3 $$ and $$ dx = dt $$. The limits of integration change as follows:

- When $$ x = 3 $$, $$ t = 0 $$.
- When $$ x = 6 $$, $$ t = 3 $$.

The integral becomes:

$$
\int_{0}^{3} \frac{\ln(t + 3)}{t^4} \, dt
$$

### Step 2: Analyze the Integrand Near $$ t = 0 $$

As $$ t $$ approaches 0, the integrand behaves like:

$$
\frac{\ln(t + 3)}{t^4} \approx \frac{\ln 3}{t^4}
$$

This approximation is because $$ \ln(t + 3) $$ approaches $$ \ln 3 $$ as $$ t \to 0 $$.

### Step 3: Determine the Convergence

Consider the integral near $$ t = 0 $$:

$$
\int_{0}^{\epsilon} \frac{\ln 3}{t^4} \, dt = \ln 3 \int_{0}^{\epsilon} t^{-4} \, dt
$$

Evaluate the integral:

$$
\ln 3 \left[ \frac{-1}{3 t^3} \right]_{0}^{\epsilon} = \ln 3 \left( \frac{-1}{3 \epsilon^3} - \left( -\infty \right) \right) = +\infty
$$

Since the integral diverges to $$ +\infty $$ as $$ t \to 0 $$, the original integral does not converge.

### Conclusion

The integral

$$
\int_{3}^{6} \frac{\ln x}{(x - 3)^4} \, dx
$$

**diverges to positive infinity**.

**Final Answer:** The integral does not converge; its value is unbounded (it diverges to infinity).
The integral does not converge; it diverges to infinity.

\( \int _ { 3 } ^ { 6 } \frac { \ln x } { ( x - 3 ) ^ { 4 } } d x = \)

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