`int(x^(2))/((x^(2)+6x-3))dx`

Explanation

To solve the integral \( \int \frac{x^2}{x^2 + 6x - 3} \, dx \), we will follow these steps:

### Step 1: Polynomial Long Division
We start by performing polynomial long division on \( \frac{x^2}{x^2 + 6x - 3} \).

1. Divide \( x^2 \) by \( x^2 + 6x - 3 \):
- The quotient is \( 1 \).
- Multiply \( 1 \) by \( (x^2 + 6x - 3) \) to get \( x^2 + 6x - 3 \).
- Subtract this from \( x^2 \):
\[ x^2 - (x^2 + 6x - 3) = -6x + 3 \]

So, we can rewrite the integral as:
\[ \int \left( 1 - \frac{6x - 3}{x^2 + 6x - 3} \right) \, dx \]

### Step 2: Split the Integral
Now we can split the integral into two parts:
\[ \int 1 \, dx - \int \frac{6x - 3}{x^2 + 6x - 3} \, dx \]

### Step 3: Integrate the First Part
The integral of \( 1 \) is straightforward:
\[ \int 1 \, dx = x \]

### Step 4: Integrate the Second Part
Now we focus on the second integral:
\[ \int \frac{6x - 3}{x^2 + 6x - 3} \, dx \]

To integrate this, we can use the substitution method. Let:
\[ u = x^2 + 6x - 3 \quad \Rightarrow \quad du = (2x + 6) \, dx \]
We can express \( dx \) in terms of \( du \):
\[ dx = \frac{du}{2x + 6} \]

Now, we need to express \( 6x - 3 \) in terms of \( u \):
\[ 6x - 3 = 6\left(\frac{u - 6x + 3}{2}\right) - 3 = \frac{6u - 36 + 18}{2} = \frac{6u - 18}{2} = 3u - 9 \]

### Step 5: Substitute and Simplify
Now substituting back into the integral:
\[ \int \frac{6x - 3}{x^2 + 6x - 3} \, dx = \int \frac{3u - 9}{u} \cdot \frac{du}{2x + 6} \]

This simplifies to:
\[ \int \left(3 - \frac{9}{u}\right) \cdot \frac{du}{2x + 6} \]

### Step 6: Integrate
Now we can integrate:
\[ \int 3 \, du - \int \frac{9}{u} \, du = 3u - 9 \ln |u| + C \]

### Step 7: Substitute Back
Substituting \( u \) back in:
\[ 3(x^2 + 6x - 3) - 9 \ln |x^2 + 6x - 3| + C \]

### Final Answer
Combining everything, we have:
\[ \int \frac{x^2}{x^2 + 6x - 3} \, dx = x + 3(x^2 + 6x - 3) - 9 \ln |x^2 + 6x - 3| + C \]