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

Answer

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

### Step 1: Simplify the Integral
We start with the integral:
\[ I = \int \frac{1}{\sqrt{3x^2 + x}} \, dx \]
We can factor out a constant from the square root:
\[ I = \int \frac{1}{\sqrt{x(3x + 1)}} \, dx \]

### Step 2: Use Substitution
To simplify the integral further, we can use a substitution. Let:
\[ x = \frac{1}{6} + u \]
Then, we have:
\[ dx = du \]
Substituting this into the integral gives:
\[ I = \int \frac{1}{\sqrt{3\left(\frac{1}{6} + u\right)^2 + \left(\frac{1}{6} + u\right)}} \, du \]

### Step 3: Expand and Simplify
Now we will expand the expression inside the square root:
\[ 3\left(\frac{1}{6} + u\right)^2 + \left(\frac{1}{6} + u\right) = 3\left(\frac{1}{36} + \frac{1}{3}u + u^2\right) + \left(\frac{1}{6} + u\right) \]
This simplifies to:
\[ \frac{1}{12} + u + 3u^2 \]

### Step 4: Rewrite the Integral
The integral now becomes:
\[ I = \int \frac{1}{\sqrt{3u^2 + u + \frac{1}{12}}} \, du \]

### Step 5: Complete the Square
To simplify the expression under the square root, we complete the square:
\[ 3u^2 + u + \frac{1}{12} = 3\left(u^2 + \frac{1}{3}u + \frac{1}{36}\right) = 3\left(u + \frac{1}{6}\right)^2 - \frac{1}{12} \]

### Step 6: Substitute Back
Substituting this back into the integral gives:
\[ I = \int \frac{1}{\sqrt{3\left(u + \frac{1}{6}\right)^2 - \frac{1}{12}}} \, du \]

### Step 7: Use the Integral Formula
Using the formula for the integral of the form \( \int \frac{dx}{\sqrt{x^2 - a^2}} = \ln |x + \sqrt{x^2 - a^2}| + C \), we can solve the integral.

### Step 8: Final Answer
After performing the integration and substituting back for \( x \), we find:
\[ I = \frac{1}{\sqrt{3}} \ln \left| x + \frac{1}{6} + \sqrt{3\left(x + \frac{1}{6}\right)^2 - \frac{1}{12}} \right| + C \]