Evaluate the following integral using integration by parts.

Integral from nothing to nothing 6 t e Superscript t Baseline dt

Let's evaluate the given integral using integration by parts.

The expression you're working with is \( \int 6t e^t \, dt \). Since no bounds are provided, we'll solve the **indefinite integral**.

### Integration by Parts Formula

The formula for integration by parts is:

\[
\int u \, dv = uv - \int v \, du
\]

### Step 1: Choose \( u \) and \( dv \)

To apply integration by parts, we first need to identify \( u \) and \( dv \). In general, we choose \( u \) to be the part that simplifies when differentiated, and \( dv \) to be the part that can be easily integrated.

Let:
- \( u = 6t \) (which simplifies upon differentiation),
- \( dv = e^t \, dt \) (which is easily integrated).

### Step 2: Differentiate \( u \) and Integrate \( dv \)

Now, we differentiate \( u \) and integrate \( dv \):

- \( du = 6 \, dt \),
- \( v = e^t \).

### Step 3: Apply the Integration by Parts Formula

Using the formula \( \int u \, dv = uv - \int v \, du \), we can now substitute the values:

\[
\int 6t e^t \, dt = 6t \cdot e^t - \int e^t \cdot 6 \, dt
\]

### Step 4: Simplify the Remaining Integral

The remaining integral is:

\[
\int 6 e^t \, dt
\]

The integral of \( e^t \) is simply \( e^t \), so:

\[
\int 6 e^t \, dt = 6 e^t
\]

### Step 5: Final Expression

Substitute this result back into the equation:

\[
\int 6t e^t \, dt = 6t e^t - 6 e^t + C
\]

Thus, the indefinite integral of \( 6t e^t \) is:

\[
\int 6t e^t \, dt = 6 e^t (t - 1) + C
\]

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

Would you like me to clarify any steps or perform additional operations, such as evaluating a definite integral if bounds are given?

### Related Questions:
1. How do you evaluate definite integrals using integration by parts?
2. Can integration by parts be applied to trigonometric functions?
3. What is the intuition behind the choice of \( u \) and \( dv \) in integration by parts?
4. How would this problem change if the exponential term were more complex (e.g., \( e^{2t} \))?
5. How is integration by parts related to the product rule for differentiation?

### Tip:
When using integration by parts, always choose \( u \) to be a function that simplifies when differentiated!

Learn how to evaluate the integral of 6t e^t using integration by parts with a step-by-step solution. This problem covers integration techniques, specifically integration by parts and handling exponential functions in calculus. Suitable for advanced high school and college-level students.

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