Evaluate: `int_(2)^(6)(2x+3)dx`.

Answer

To evaluate the integral \( \int_{2}^{6} (2x + 3) \, dx \), we will follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ \int_{2}^{6} (2x + 3) \, dx \]

### Step 2: Integrate the function
To integrate \( 2x + 3 \), we can break it down into two parts:
1. The integral of \( 2x \)
2. The integral of \( 3 \)

The integral of \( 2x \) is:
\[ \int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2 \]

The integral of \( 3 \) is:
\[ \int 3 \, dx = 3x \]

Thus, the integral of \( 2x + 3 \) is:
\[ \int (2x + 3) \, dx = x^2 + 3x + C \]
where \( C \) is the constant of integration.

### Step 3: Evaluate the definite integral
Now we will evaluate the definite integral from \( 2 \) to \( 6 \):
\[ \left[ x^2 + 3x \right]_{2}^{6} \]

Calculating at the upper limit \( x = 6 \):
\[ 6^2 + 3 \cdot 6 = 36 + 18 = 54 \]

Calculating at the lower limit \( x = 2 \):
\[ 2^2 + 3 \cdot 2 = 4 + 6 = 10 \]

### Step 4: Subtract the lower limit from the upper limit
Now we subtract the value at the lower limit from the value at the upper limit:
\[ 54 - 10 = 44 \]

### Final Answer
Thus, the value of the integral \( \int_{2}^{6} (2x + 3) \, dx \) is:
\[ \boxed{44} \]