`int(1)/(x^(2)+4x+13)dx` is equal to

Options

A. `log(x^(2)+4x+13)+C`
B. `(1)/(3)tan^(-1)((x+2)/(3))`
C. `log(2x+4)+C`
D. `(2x+4)/((x^(2)+4x+13)^(3))+C`

Explanation

To solve the integral \( \int \frac{1}{x^2 + 4x + 13} \, dx \), we will follow these steps:

### Step 1: Complete the square for the quadratic expression
The first step is to rewrite the quadratic expression in the denominator in a more manageable form. We can complete the square for \( x^2 + 4x + 13 \).

\[ x^2 + 4x + 13 = (x^2 + 4x + 4) + 9 = (x + 2)^2 + 9 \]

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

\[ \int \frac{1}{(x + 2)^2 + 3^2} \, dx \]

### Step 3: Use the formula for integration
We can use the formula for the integral of the form \( \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \).

In our case, \( a = 3 \) and we need to adjust for the \( (x + 2) \) term:

\[ \int \frac{1}{(x + 2)^2 + 3^2} \, dx = \frac{1}{3} \tan^{-1} \left( \frac{x + 2}{3} \right) + C \]

### Final Answer
Thus, the integral evaluates to:

\[ \int \frac{1}{x^2 + 4x + 13} \, dx = \frac{1}{3} \tan^{-1} \left( \frac{x + 2}{3} \right) + C \]