`intsqrt(5-2x+x^(2))dx`

Explanation

To solve the integral \(\int \sqrt{5 - 2x + x^2} \, dx\), we can follow these steps:

### Step 1: Rewrite the expression under the square root
We start by rewriting the expression \(5 - 2x + x^2\) in a more manageable form. Notice that it can be rearranged as:
\[ x^2 - 2x + 5 = (x - 1)^2 + 4 \]
This is because:
\[ (x - 1)^2 = x^2 - 2x + 1 \]
Thus,
\[ (x - 1)^2 + 4 = x^2 - 2x + 1 + 4 = x^2 - 2x + 5 \]

### Step 2: Substitute into the integral
Now, we can substitute this back into the integral:
\[ \int \sqrt{(x - 1)^2 + 4} \, dx \]

### Step 3: Use a trigonometric substitution
We can use a trigonometric substitution to simplify the integral. Let:
\[ x - 1 = 2\tan(\theta) \quad \Rightarrow \quad dx = 2\sec^2(\theta) \, d\theta \]
The expression under the square root becomes:
\[ \sqrt{(2\tan(\theta))^2 + 4} = \sqrt{4\tan^2(\theta) + 4} = \sqrt{4(\tan^2(\theta) + 1)} = 2\sec(\theta) \]

### Step 4: Substitute into the integral
Now substitute these into the integral:
\[ \int \sqrt{(x - 1)^2 + 4} \, dx = \int 2\sec(\theta) \cdot 2\sec^2(\theta) \, d\theta = 4 \int \sec^3(\theta) \, d\theta \]

### Step 5: Integrate \(\sec^3(\theta)\)
The integral of \(\sec^3(\theta)\) is known:
\[ \int \sec^3(\theta) \, d\theta = \frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln | \sec(\theta) + \tan(\theta) | + C \]
Thus,
\[ 4 \int \sec^3(\theta) \, d\theta = 2 \sec(\theta) \tan(\theta) + 2 \ln | \sec(\theta) + \tan(\theta) | + C \]

### Step 6: Substitute back to \(x\)
Now we need to substitute back \(\theta\) in terms of \(x\):
- From \(x - 1 = 2\tan(\theta)\), we have \(\tan(\theta) = \frac{x - 1}{2}\).
- Therefore, \(\sec(\theta) = \sqrt{1 + \tan^2(\theta)} = \sqrt{1 + \left(\frac{x - 1}{2}\right)^2} = \sqrt{\frac{(x - 1)^2 + 4}{4}} = \frac{\sqrt{(x - 1)^2 + 4}}{2}\).

Thus,
\[ \tan(\theta) = \frac{x - 1}{2} \quad \text{and} \quad \sec(\theta) = \frac{\sqrt{(x - 1)^2 + 4}}{2} \]

### Final Result
Substituting these back into our integral, we have:
\[ \int \sqrt{5 - 2x + x^2} \, dx = 2 \cdot \frac{\sqrt{(x - 1)^2 + 4}}{2} \cdot \frac{x - 1}{2} + 2 \ln \left| \frac{\sqrt{(x - 1)^2 + 4}}{2} + \frac{x - 1}{2} \right| + C \]
Simplifying gives:
\[ = (x - 1) \sqrt{(x - 1)^2 + 4} + 2 \ln \left| \sqrt{(x - 1)^2 + 4} + (x - 1) \right| + C \]