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

Answer

To evaluate the integral

\[ I = \int \frac{2x + 5}{\sqrt{x^2 + 2x + 5}} \, dx, \]

we can break it down into simpler parts.

### Step 1: Rewrite the integrand
We can rewrite \(5\) as \(2 + 3\):

\[ I = \int \frac{2x + 2 + 3}{\sqrt{x^2 + 2x + 5}} \, dx = \int \frac{2x + 2}{\sqrt{x^2 + 2x + 5}} \, dx + \int \frac{3}{\sqrt{x^2 + 2x + 5}} \, dx. \]

### Step 2: Simplify the first integral
Now, we can factor out \(2\) from the first part:

\[ I_1 = \int \frac{2(x + 1)}{\sqrt{x^2 + 2x + 5}} \, dx. \]

### Step 3: Substitution
Let \(t = x^2 + 2x + 5\). Then, differentiate both sides:

\[ dt = (2x + 2) \, dx \implies dx = \frac{dt}{2(x + 1)}. \]

Substituting into \(I_1\):

\[ I_1 = \int \frac{2(x + 1)}{\sqrt{t}} \cdot \frac{dt}{2(x + 1)} = \int \frac{1}{\sqrt{t}} \, dt. \]

### Step 4: Integrate
The integral becomes:

\[ I_1 = \int t^{-1/2} \, dt = 2\sqrt{t} + C_1. \]

### Step 5: Substitute back for \(t\)
Substituting back \(t = x^2 + 2x + 5\):

\[ I_1 = 2\sqrt{x^2 + 2x + 5}. \]

### Step 6: Evaluate the second integral
Now consider the second part:

\[ I_2 = \int \frac{3}{\sqrt{x^2 + 2x + 5}} \, dx. \]

### Step 7: Complete the square
We can rewrite \(x^2 + 2x + 5\) as:

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

### Step 8: Use a trigonometric substitution
Using the formula for the integral of the form \(\int \frac{1}{\sqrt{x^2 + a^2}} \, dx\):

\[ I_2 = 3 \int \frac{1}{\sqrt{(x + 1)^2 + 2^2}} \, dx = 3 \left( \log |x + 1 + \sqrt{(x + 1)^2 + 4}| + C_2 \right). \]

### Step 9: Combine results
Combining \(I_1\) and \(I_2\):

\[ I = 2\sqrt{x^2 + 2x + 5} + 3 \log |x + 1 + \sqrt{(x + 1)^2 + 4}| + C. \]

### Final Answer
Thus, the final result of the integral is:

\[ I = 2\sqrt{x^2 + 2x + 5} + 3 \log |x + 1 + \sqrt{(x + 1)^2 + 4}| + C. \]