To solve the integral of $\ln(2x)$, we can use integration by parts. Here's the step-by-step process:

We want to compute:

$\int \ln(2x) \, dx$

Step 1: Recall the formula for integration by parts

The formula for integration by parts is:

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

For this integral, we choose:

• $u = \ln(2x)$ (since differentiating logarithms simplifies the expression)
• $dv = dx$

Step 2: Differentiate and integrate

Now, we compute $du$ and $v$:

• $du = \frac{d}{dx} \ln(2x) = \frac{1}{x} \, dx$
• $v = \int dx = x$

Step 3: Apply the integration by parts formula

Substituting into the formula:

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

Simplifying the remaining integral:

$\int \ln(2x) \, dx = x \ln(2x) - \int 1 \, dx$

Step 4: Final computation

The integral of 1 is just $x$, so:

$\int \ln(2x) \, dx = x \ln(2x) - x + C$

where $C$ is the constant of integration.

Thus, the final answer is:

$\boxed{\int \ln(2x) \, dx = x \ln(2x) - x + C}$

Would you like more details on any step or have any questions?

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Here are 5 related questions to expand on this topic:

1. What is the integration by parts method and when is it useful?
2. How do you integrate functions involving logarithms in general?
3. What are other examples of integrals where substitution may be used instead of parts?
4. Can you integrate $\ln(3x)$ and see the difference in steps?
5. How does the chain rule apply to logarithmic differentiation?

Tip: Always check if integration by parts can simplify a problem when dealing with a product of two functions.