Evaluate `int sqrt(x^(2)+2x+5)dx`

Explanation

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

### Step 1: Simplify the expression under the square root
We start with the expression \( x^2 + 2x + 5 \). We can complete the square for the quadratic expression:

\[ x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x + 1)^2 + 4 \]

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

\[ \int \sqrt{x^2 + 2x + 5} \, dx = \int \sqrt{(x + 1)^2 + 4} \, dx \]

### Step 3: Use a substitution
Let \( u = x + 1 \), then \( du = dx \). The integral becomes:

\[ \int \sqrt{u^2 + 4} \, du \]

### Step 4: Use the formula for integration
The integral \( \int \sqrt{u^2 + a^2} \, du \) can be evaluated using the formula:

\[ \int \sqrt{u^2 + a^2} \, du = \frac{u}{2} \sqrt{u^2 + a^2} + \frac{a^2}{2} \ln \left| u + \sqrt{u^2 + a^2} \right| + C \]

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

\[ \int \sqrt{u^2 + 4} \, du = \frac{u}{2} \sqrt{u^2 + 4} + \frac{4}{2} \ln \left| u + \sqrt{u^2 + 4} \right| + C \]

### Step 5: Substitute back for \( u \)
Now, substituting back \( u = x + 1 \):

\[ = \frac{x + 1}{2} \sqrt{(x + 1)^2 + 4} + 2 \ln \left| (x + 1) + \sqrt{(x + 1)^2 + 4} \right| + C \]

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

\[ \int \sqrt{x^2 + 2x + 5} \, dx = \frac{x + 1}{2} \sqrt{(x + 1)^2 + 4} + 2 \ln \left| (x + 1) + \sqrt{(x + 1)^2 + 4} \right| + C \]