Evaluate the following integrals:
`int frac{3x-1}{sqrt(x^2+9)}dx`

Answer

To evaluate the integral \( \int \frac{3x-1}{\sqrt{x^2+9}} \, dx \), we can follow these steps:

### Step 1: Split the Integral
We can separate the integral into two parts:
\[ \int \frac{3x-1}{\sqrt{x^2+9}} \, dx = \int \frac{3x}{\sqrt{x^2+9}} \, dx - \int \frac{1}{\sqrt{x^2+9}} \, dx \]

### Step 2: Evaluate the First Integral
For the first integral \( \int \frac{3x}{\sqrt{x^2+9}} \, dx \), we can use substitution. Let:
\[ t = x^2 + 9 \implies dt = 2x \, dx \implies dx = \frac{dt}{2x} \]
Now, substituting \( x = \sqrt{t - 9} \):
\[ \int \frac{3x}{\sqrt{x^2+9}} \, dx = \int \frac{3\sqrt{t-9}}{\sqrt{t}} \cdot \frac{dt}{2\sqrt{t-9}} = \frac{3}{2} \int dt = \frac{3}{2} t + C_1 = \frac{3}{2} (x^2 + 9) + C_1 \]

### Step 3: Evaluate the Second Integral
Now, we evaluate the second integral \( \int \frac{1}{\sqrt{x^2+9}} \, dx \). This is a standard integral:
\[ \int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \log |x + \sqrt{x^2 + a^2}| + C \]
Here, \( a = 3 \):
\[ \int \frac{1}{\sqrt{x^2 + 9}} \, dx = \log |x + \sqrt{x^2 + 9}| + C_2 \]

### Step 4: Combine the Results
Now we combine the results from Step 2 and Step 3:
\[ \int \frac{3x-1}{\sqrt{x^2+9}} \, dx = \frac{3}{2} (x^2 + 9) - \log |x + \sqrt{x^2 + 9}| + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{3x-1}{\sqrt{x^2+9}} \, dx = \frac{3}{2} x^2 + \frac{27}{2} - \log |x + \sqrt{x^2 + 9}| + C \]