`int(dx)/((x^(2)+4x+8))`

Answer

To solve the integral \(\int \frac{dx}{x^2 + 4x + 8}\), we will follow these steps:

### Step 1: Rewrite the Denominator
First, we need to rewrite the quadratic expression in the denominator \(x^2 + 4x + 8\) in a more manageable form. We can complete the square.

\[ x^2 + 4x + 8 = (x^2 + 4x + 4) + 4 = (x + 2)^2 + 4 \]

### Step 2: Substitute into the Integral
Now, we can substitute this back into the integral:

\[ \int \frac{dx}{(x + 2)^2 + 2^2} \]

### Step 3: Identify the Integral Formula
We recognize that this integral matches the standard form:

\[ \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C \]

In our case, \(a = 2\) and we have \(x\) replaced by \((x + 2)\).

### Step 4: Apply the Integral Formula
Using the formula, we have:

\[ \int \frac{dx}{(x + 2)^2 + 2^2} = \frac{1}{2} \tan^{-1} \left(\frac{x + 2}{2}\right) + C \]

### Final Answer
Thus, the final answer for the integral is:

\[ \int \frac{dx}{x^2 + 4x + 8} = \frac{1}{2} \tan^{-1} \left(\frac{x + 2}{2}\right) + C \]

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