Evaluate: `int (dx)/(x^2-6x+13)`

Answer

To evaluate the integral \(\int \frac{dx}{x^2 - 6x + 13}\), we can follow these steps:

### Step 1: Complete the square in the denominator
The expression in the denominator is \(x^2 - 6x + 13\). We can complete the square for the quadratic expression.

\[ x^2 - 6x + 13 = (x^2 - 6x + 9) + 4 = (x - 3)^2 + 4 \]

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

\[ \int \frac{dx}{(x - 3)^2 + 4} \]

### Step 3: Use a substitution
Let \(t = x - 3\), then \(dx = dt\). The integral becomes:

\[ \int \frac{dt}{t^2 + 4} \]

### Step 4: Recognize the form of the integral
The integral \(\int \frac{dt}{t^2 + a^2}\) has a known solution:

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

In our case, \(a^2 = 4\) implies \(a = 2\). Therefore, we have:

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

### Step 5: Substitute back for \(t\)
Now we substitute back \(t = x - 3\):

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

### Final Answer
Thus, the evaluated integral is:

\[ \int \frac{dx}{x^2 - 6x + 13} = \frac{1}{2} \tan^{-1} \left( \frac{x - 3}{2} \right) + C \]