To integrate $x \log(1 + 2x)$, we can use integration by parts. We set it up as follows:

Step 1: Apply integration by parts formula

The integration by parts formula is:

$\int u \, dv = uv - \int v \, du$

Let:

• $u = \log(1 + 2x)$, so $du = \frac{2}{1 + 2x} dx$
• $dv = x \, dx$, so $v = \frac{x^2}{2}$

Step 2: Substitute into the formula

Now, substitute $u$, $v$, $du$, and $dv$ into the formula:

$\int x \log(1 + 2x) \, dx = \frac{x^2}{2} \log(1 + 2x) - \int \frac{x^2}{2} \cdot \frac{2}{1 + 2x} \, dx$

Simplifying the integral on the right-hand side:

$= \frac{x^2}{2} \log(1 + 2x) - \int \frac{x^2}{1 + 2x} \, dx$

Step 3: Solve the remaining integral

To solve $\int \frac{x^2}{1 + 2x} \, dx$, we use polynomial long division. We divide $x^2$ by $1 + 2x$:

$\frac{x^2}{1 + 2x} = \frac{1}{4}(2x - 1) + \frac{1}{4(1 + 2x)}$

Thus:

$\int \frac{x^2}{1 + 2x} \, dx = \frac{1}{4} \int (2x - 1) \, dx + \frac{1}{4} \int \frac{1}{1 + 2x} \, dx$

Step 4: Solve each integral

• $\frac{1}{4} \int (2x - 1) \, dx = \frac{1}{4} \left( x^2 - x \right)$
• $\frac{1}{4} \int \frac{1}{1 + 2x} \, dx = \frac{1}{8} \log(1 + 2x)$

Step 5: Final answer

Putting everything together:

$\int x \log(1 + 2x) \, dx = \frac{x^2}{2} \log(1 + 2x) - \frac{1}{4} \left( x^2 - x \right) - \frac{1}{8} \log(1 + 2x) + C$

This is the final result.

Would you like more details on any step? Here are 5 related questions for practice:

1. How do you integrate by parts for the function $x e^x$?
2. What is the general strategy for polynomial long division?
3. Can you derive the formula for integration by parts from the product rule?
4. How do you handle integrals involving logarithmic functions in general?
5. What are some common techniques for integrating rational functions?

Tip: Whenever facing an integral involving a product of two different functions, integration by parts is often a good first approach.