To integrate the function \( x \cdot 2^x \), we can use integration by parts. Here’s the step-by-step process:

### Step 1: Set up integration by parts
The formula for integration by parts is:
\[
\int u \, dv = u \cdot v - \int v \, du
\]
In this case, we need to select \( u \) and \( dv \). We'll let:
- \( u = x \) (since it's simpler to differentiate)
- \( dv = 2^x \, dx \)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now, find \( du \) and \( v \):
- \( u = x \), so \( du = dx \)
- \( dv = 2^x \, dx \), so \( v = \frac{2^x}{\ln 2} \) (using the rule for the integral of \( a^x \), which is \( \frac{a^x}{\ln a} \)).

### Step 3: Apply the integration by parts formula
Now substitute these into the formula:
\[
\int x \cdot 2^x \, dx = u \cdot v - \int v \cdot du
\]
\[
= x \cdot \frac{2^x}{\ln 2} - \int \frac{2^x}{\ln 2} \, dx
\]

### Step 4: Simplify the remaining integral
The remaining integral is:
\[
\int \frac{2^x}{\ln 2} \, dx = \frac{2^x}{(\ln 2)^2}
\]

### Step 5: Combine the results
Now, combining everything, the solution to the integral is:
\[
\int x \cdot 2^x \, dx = \frac{x \cdot 2^x}{\ln 2} - \frac{2^x}{(\ln 2)^2} + C
\]
where \( C \) is the constant of integration.

Would you like more details on this technique, or clarification on any step?

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Here are 5 related questions to explore this topic further:
1. How would you apply integration by parts to \( x \cdot e^x \)?
2. What are the conditions for using integration by parts effectively?
3. How does the integral of \( a^x \) differ from \( e^x \)?
4. Can integration by parts be applied multiple times to a single integral?
5. What happens when you integrate a product of polynomials and exponentials?

**Tip**: When using integration by parts, choose \( u \) to simplify upon differentiation and \( dv \) to be easy to integrate.

Learn how to solve the integral of x * 2^x using the method of integration by parts. This detailed solution breaks down the process step-by-step, covering key concepts such as exponential functions and integration techniques. Suitable for undergraduate or advanced high school calculus students.

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