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

Answer

To evaluate the integral

\[ I = \int \frac{3x - 1}{\sqrt{x^2 + 9}} \, dx, \]

we can break it down into two separate integrals:

\[ I = \int \frac{3x}{\sqrt{x^2 + 9}} \, dx - \int \frac{1}{\sqrt{x^2 + 9}} \, dx. \]

### Step 1: Evaluate the first integral

For the first integral, we can use the substitution \( t = x^2 + 9 \). Then, differentiating both sides gives us:

\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x}. \]

Now, substituting \( t \) into the integral, we have:

\[ \int \frac{3x}{\sqrt{x^2 + 9}} \, dx = \int \frac{3x}{\sqrt{t}} \cdot \frac{dt}{2x} = \frac{3}{2} \int \frac{1}{\sqrt{t}} \, dt. \]

The integral of \( \frac{1}{\sqrt{t}} \) is \( 2\sqrt{t} \), so we get:

\[ \frac{3}{2} \cdot 2\sqrt{t} = 3\sqrt{t}. \]

Substituting back for \( t \):

\[ 3\sqrt{x^2 + 9}. \]

### Step 2: Evaluate the second integral

Now we evaluate the second integral:

\[ \int \frac{1}{\sqrt{x^2 + 9}} \, dx. \]

Using the formula

\[ \int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \ln |x + \sqrt{x^2 + a^2}| + C, \]

with \( a = 3 \):

\[ \int \frac{1}{\sqrt{x^2 + 9}} \, dx = \ln |x + \sqrt{x^2 + 9}| + C. \]

### Step 3: Combine results

Now, combining both parts, we have:

\[ I = 3\sqrt{x^2 + 9} - \ln |x + \sqrt{x^2 + 9}| + C. \]

### Final Answer

Thus, the evaluated integral is:

\[ \int \frac{3x - 1}{\sqrt{x^2 + 9}} \, dx = 3\sqrt{x^2 + 9} - \ln |x + \sqrt{x^2 + 9}| + C. \]

---