`I= int - (dx)/( sqrt( x^(2) + 6x + 1) )`

Answer

To solve the integral \( I = -\int \frac{dx}{\sqrt{x^2 + 6x + 1}} \), we will follow these steps:

### Step 1: Simplify the expression under the square root
First, we need to complete the square for the expression \( x^2 + 6x + 1 \).

\[ x^2 + 6x + 1 = (x^2 + 6x + 9) - 9 + 1 = (x + 3)^2 - 8 \]

### Step 2: Rewrite the integral
Now we can rewrite the integral using the completed square:

\[ I = -\int \frac{dx}{\sqrt{(x + 3)^2 - 8}} \]

### Step 3: Use a trigonometric substitution
To evaluate this integral, we can use the substitution \( x + 3 = 2\sqrt{2} \sec(\theta) \). Then, \( dx = 2\sqrt{2} \sec(\theta) \tan(\theta) d\theta \).

### Step 4: Substitute into the integral
Substituting \( x + 3 = 2\sqrt{2} \sec(\theta) \) into the integral gives:

\[ I = -\int \frac{2\sqrt{2} \sec(\theta) \tan(\theta) d\theta}{\sqrt{(2\sqrt{2} \sec(\theta))^2 - 8}} \]

The expression under the square root simplifies to:

\[ \sqrt{(2\sqrt{2})^2 \sec^2(\theta) - 8} = \sqrt{8 \sec^2(\theta) - 8} = \sqrt{8(\sec^2(\theta) - 1)} = \sqrt{8 \tan^2(\theta)} = 2\sqrt{2} \tan(\theta) \]

Thus, the integral becomes:

\[ I = -\int \frac{2\sqrt{2} \sec(\theta) \tan(\theta) d\theta}{2\sqrt{2} \tan(\theta)} = -\int \sec(\theta) d\theta \]

### Step 5: Integrate
The integral of \( \sec(\theta) \) is:

\[ -\int \sec(\theta) d\theta = -\ln |\sec(\theta) + \tan(\theta)| + C \]

### Step 6: Back substitute
Now we need to back substitute for \( \theta \). Recall that:

\[ \tan(\theta) = \frac{\sqrt{(x + 3)^2 - 8}}{2\sqrt{2}} \quad \text{and} \quad \sec(\theta) = \frac{x + 3}{2\sqrt{2}} \]

Thus, we have:

\[ I = -\ln \left| \frac{x + 3}{2\sqrt{2}} + \frac{\sqrt{(x + 3)^2 - 8}}{2\sqrt{2}} \right| + C \]

### Final Answer
The final answer can be simplified to:

\[ I = -\ln \left| (x + 3) + \sqrt{(x + 3)^2 - 8} \right| + C \]