`int(2x+5)/(sqrt(x^(2)+3x+1))dx`

Answer

To solve the integral \(\int \frac{2x + 5}{\sqrt{x^2 + 3x + 1}} \, dx\), we can follow these steps:

### Step 1: Rewrite the Numerator
We want to express the numerator \(2x + 5\) in a way that relates it to the derivative of the denominator. We can write:
\[ 2x + 5 = 2x + 3 + 2 = (2x + 3) + 2 \]
This allows us to separate the integral into two parts:
\[ \int \frac{2x + 5}{\sqrt{x^2 + 3x + 1}} \, dx = \int \frac{(2x + 3) + 2}{\sqrt{x^2 + 3x + 1}} \, dx \]

### Step 2: Separate the Integral
Now we can separate the integral:
\[ \int \frac{2x + 5}{\sqrt{x^2 + 3x + 1}} \, dx = \int \frac{2x + 3}{\sqrt{x^2 + 3x + 1}} \, dx + \int \frac{2}{\sqrt{x^2 + 3x + 1}} \, dx \]

### Step 3: Solve the First Integral
For the first integral \(\int \frac{2x + 3}{\sqrt{x^2 + 3x + 1}} \, dx\), we can use substitution. Let:
\[ u = x^2 + 3x + 1 \quad \Rightarrow \quad du = (2x + 3) \, dx \]
Thus, the integral becomes:
\[ \int \frac{du}{\sqrt{u}} = 2\sqrt{u} + C_1 = 2\sqrt{x^2 + 3x + 1} + C_1 \]

### Step 4: Solve the Second Integral
Now for the second integral \(\int \frac{2}{\sqrt{x^2 + 3x + 1}} \, dx\), we need to rewrite the expression under the square root. We complete the square:
\[ x^2 + 3x + 1 = (x + \frac{3}{2})^2 - \frac{5}{4} \]
Thus, we can rewrite the integral as:
\[ \int \frac{2}{\sqrt{(x + \frac{3}{2})^2 - \frac{5}{4}}} \, dx \]
This is of the form \(\int \frac{1}{\sqrt{x^2 - a^2}} \, dx\), which integrates to:
\[ 2 \log \left| x + \frac{3}{2} + \sqrt{(x + \frac{3}{2})^2 - \frac{5}{4}} \right| + C_2 \]

### Step 5: Combine the Results
Combining both parts, we have:
\[ \int \frac{2x + 5}{\sqrt{x^2 + 3x + 1}} \, dx = 2\sqrt{x^2 + 3x + 1} + 2 \log \left| x + \frac{3}{2} + \sqrt{(x + \frac{3}{2})^2 - \frac{5}{4}} \right| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{2x + 5}{\sqrt{x^2 + 3x + 1}} \, dx = 2\sqrt{x^2 + 3x + 1} + 2 \log \left| x + \frac{3}{2} + \sqrt{(x + \frac{3}{2})^2 - \frac{5}{4}} \right| + C \]