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

Answer

To solve the integral

\[ I = \int \frac{2x + 5}{\sqrt{x^2 + 3x + 1}} \, dx, \]

we can break it down into simpler parts. Let's follow the steps outlined in the video transcript.

### Step 1: Split the Integral
We can rewrite the integral by splitting the constant \(5\) into \(3 + 2\):

\[ I = \int \frac{2x + 3 + 2}{\sqrt{x^2 + 3x + 1}} \, dx = \int \frac{2x + 3}{\sqrt{x^2 + 3x + 1}} \, dx + \int \frac{2}{\sqrt{x^2 + 3x + 1}} \, dx. \]

Let’s denote the first integral as \(I_1\) and the second as \(I_2\):

\[ I = I_1 + I_2, \]

where

\[ I_1 = \int \frac{2x + 3}{\sqrt{x^2 + 3x + 1}} \, dx, \]
and
\[ I_2 = \int \frac{2}{\sqrt{x^2 + 3x + 1}} \, dx. \]

### Step 2: Solve \(I_1\)
For \(I_1\), we can use substitution. Let

\[ t = x^2 + 3x + 1. \]

Then, differentiating gives:

\[ dt = (2x + 3) \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x + 3}. \]

Substituting into \(I_1\):

\[ I_1 = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} + C_1 = 2\sqrt{x^2 + 3x + 1} + C_1. \]

### Step 3: Solve \(I_2\)
Next, we simplify \(I_2\). We need to complete the square for the expression \(x^2 + 3x + 1\):

\[ x^2 + 3x + 1 = \left(x + \frac{3}{2}\right)^2 - \frac{5}{4}. \]

Thus,

\[ I_2 = 2 \int \frac{1}{\sqrt{\left(x + \frac{3}{2}\right)^2 - \left(\frac{\sqrt{5}}{2}\right)^2}} \, dx. \]

Using the formula for the integral of the form \(\int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln |x + \sqrt{x^2 - a^2}| + C\), we have:

\[ I_2 = 2 \ln \left| x + \frac{3}{2} + \sqrt{\left(x + \frac{3}{2}\right)^2 - \frac{5}{4}} \right| + C_2. \]

### Step 4: Combine Results
Now, we combine \(I_1\) and \(I_2\):

\[ I = 2\sqrt{x^2 + 3x + 1} + 2 \ln \left| x + \frac{3}{2} + \sqrt{\left(x + \frac{3}{2}\right)^2 - \frac{5}{4}} \right| + C, \]

where \(C = C_1 + C_2\).

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

\[ I = 2\sqrt{x^2 + 3x + 1} + 2 \ln \left| x + \frac{3}{2} + \sqrt{\left(x + \frac{3}{2}\right)^2 - \frac{5}{4}} \right| + C. \]

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