Exaluate integral of (18x+24)dx over 9x^2+24x+6

To evaluate the integral 

\[
\int \frac{18x + 24}{9x^2 + 24x + 6} \, dx,
\]

we can proceed with the following steps:

### Step 1: Simplify the integral

First, observe the derivative of the quadratic function in the denominator:

\[
\frac{d}{dx}(9x^2 + 24x + 6) = 18x + 24.
\]

This is exactly the numerator of our integral. Therefore, we can use the substitution method.

### Step 2: Substitution

Let 

\[
u = 9x^2 + 24x + 6.
\]

Then, 

\[
\frac{du}{dx} = 18x + 24 \quad \Rightarrow \quad du = (18x + 24)dx.
\]

This means the integral simplifies to:

\[
\int \frac{du}{u}.
\]

### Step 3: Integrate

The integral of \(\frac{1}{u}\) with respect to \(u\) is:

\[
\int \frac{du}{u} = \ln|u| + C,
\]

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

### Step 4: Back-substitute \(u\)

Substitute back \(u = 9x^2 + 24x + 6\):

\[
\ln|9x^2 + 24x + 6| + C.
\]

### Final Answer

Thus, the integral is:

\[
\int \frac{18x + 24}{9x^2 + 24x + 6} \, dx = \ln|9x^2 + 24x + 6| + C.
\]

Would you like more details, or do you have any questions? 

Here are 8 related questions for further exploration:

1. How would you integrate \(\int \frac{4x + 5}{2x^2 + 5x + 1} \, dx\)?
2. Can you explain the substitution method in integration?
3. How would you solve the integral \(\int \frac{1}{x^2 + 1} \, dx\)?
4. What is the importance of recognizing the derivative of the denominator in an integral?
5. How would you integrate a function where the numerator is not the derivative of the denominator?
6. What is the integral of \(\int e^{2x} \, dx\)?
7. How can you solve \(\int \frac{dx}{x^2 + 1}\)?
8. What is the integral of \(\int \frac{3x^2}{x^3 + 1} \, dx\)?

**Tip:** When you see a rational function, check if the numerator is the derivative of the denominator; this simplifies the integration process significantly.

Learn how to evaluate the integral of (18x + 24)/(9x^2 + 24x + 6) dx step-by-step with detailed explanations. This problem covers integration techniques including substitution method and handling rational functions. Suitable for advanced high school students and calculus enthusiasts.

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