`int(3x-1)/(sqrt(x^(2)+9))dx`

Explanation

To solve the integral \( \int \frac{3x - 1}{\sqrt{x^2 + 9}} \, dx \), we can break it down into two separate integrals:

1. \( \int \frac{3x}{\sqrt{x^2 + 9}} \, dx \)
2. \( \int \frac{-1}{\sqrt{x^2 + 9}} \, dx \)

So, we can write:

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

### Step 1: Solve \( \int \frac{3x}{\sqrt{x^2 + 9}} \, dx \)

To solve this integral, we can use the substitution method. Let:

\[ u = x^2 + 9 \quad \Rightarrow \quad du = 2x \, dx \quad \Rightarrow \quad dx = \frac{du}{2x} \]

Now, substituting \( x = \sqrt{u - 9} \) into the integral:

\[ \int \frac{3x}{\sqrt{x^2 + 9}} \, dx = \int \frac{3x}{\sqrt{u}} \cdot \frac{du}{2x} = \frac{3}{2} \int \frac{1}{\sqrt{u}} \, du \]

Now, integrating:

\[ \frac{3}{2} \int u^{-1/2} \, du = \frac{3}{2} \cdot 2u^{1/2} + C = 3\sqrt{u} + C = 3\sqrt{x^2 + 9} + C \]

### Step 2: Solve \( \int \frac{-1}{\sqrt{x^2 + 9}} \, dx \)

This integral can be solved using a standard integral formula:

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

Here, \( a = 3 \):

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

### Final Solution

Combining both parts, we have:

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