Evaluate the following integrals:
`int frac{2x+3}{x^2+3x}dx`

Answer

To evaluate the integral

\[ \int \frac{2x + 3}{x^2 + 3x} \, dx, \]

we can follow these steps:

### Step 1: Identify the Denominator and its Derivative
The denominator is \( x^2 + 3x \). Let's denote it as \( f(x) = x^2 + 3x \).

### Step 2: Differentiate the Denominator
Now, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}(x^2 + 3x) = 2x + 3. \]

### Step 3: Relate the Numerator to the Derivative
Notice that the numerator \( 2x + 3 \) is exactly the derivative \( f'(x) \). This gives us a direct relationship:

\[ \frac{2x + 3}{x^2 + 3x} = \frac{f'(x)}{f(x)}. \]

### Step 4: Apply the Logarithmic Integration Rule
Using the property of integrals, we know that:

\[ \int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C, \]

where \( C \) is the constant of integration.

### Step 5: Substitute Back the Function
Substituting back \( f(x) = x^2 + 3x \), we have:

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

### Final Answer
Thus, the integral evaluates to:

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

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