Evaluate `int(1)/(sqrt(x^(2)-2x+3))dx`

Answer

To evaluate the integral \( \int \frac{1}{\sqrt{x^2 - 2x + 3}} \, dx \), we can follow these steps:

### Step 1: Complete the square
First, we need to rewrite the expression under the square root in a more manageable form. The expression \( x^2 - 2x + 3 \) can be rewritten by completing the square.

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

### Step 2: Substitute
Now, we can substitute \( t = x - 1 \). Therefore, \( dx = dt \).

The integral now becomes:

\[ \int \frac{1}{\sqrt{(x - 1)^2 + 2}} \, dx = \int \frac{1}{\sqrt{t^2 + 2}} \, dt \]

### Step 3: Use a standard integral formula
We can use the standard integral formula:

\[ \int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \ln |x + \sqrt{x^2 + a^2}| + C \]

In our case, \( a^2 = 2 \), so \( a = \sqrt{2} \). Thus, we have:

\[ \int \frac{1}{\sqrt{t^2 + 2}} \, dt = \ln |t + \sqrt{t^2 + 2}| + C \]

### Step 4: Substitute back
Now we substitute back \( t = x - 1 \):

\[ \ln |(x - 1) + \sqrt{(x - 1)^2 + 2}| + C \]

### Step 5: Simplify the expression
Now we simplify the expression inside the logarithm:

\[ \sqrt{(x - 1)^2 + 2} = \sqrt{x^2 - 2x + 3} \]

Thus, the final answer is:

\[ \ln |(x - 1) + \sqrt{x^2 - 2x + 3}| + C \]

### Final Answer
\[ \int \frac{1}{\sqrt{x^2 - 2x + 3}} \, dx = \ln |(x - 1) + \sqrt{x^2 - 2x + 3}| + C \]

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