9–40. Integration by parts Evaluate the following integrals using integration by parts.
17. ∫ x · 3x dx

Verified step by step guidance

1. Identify the integral to solve: \(\int x \cdot 3^x \, dx\).
2. Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\).
3. Choose \(u\) and \(dv\) wisely. Let \(u = x\) (which simplifies when differentiated) and \(dv = 3^x \, dx\) (which can be integrated).
4. Compute \(du\) by differentiating \(u\): \(du = dx\). Compute \(v\) by integrating \(dv\): \(v = \int 3^x \, dx\); remember that \(\int a^x \, dx = \frac{a^x}{\ln(a)} + C\) for \(a > 0\), \(a \neq 1\).
5. Apply the integration by parts formula: \(\int x \cdot 3^x \, dx = u v - \int v \, du = x \cdot v - \int v \, dx\). Then simplify and evaluate the remaining integral.

Key Concepts

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It transforms the integral of a product of functions into simpler integrals using the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely is crucial to simplify the problem.

Choosing u and dv

In integration by parts, selecting u (a function to differentiate) and dv (a function to integrate) affects the ease of solving the integral. Typically, u is chosen as a function that simplifies when differentiated, and dv is chosen as a function that is easy to integrate.

Integrating Exponential Functions

When integrating expressions involving exponential functions like 3^x, recall that the integral of a^x is (a^x)/(ln a) + C for a > 0, a ≠ 1. This knowledge helps in computing dv or v when the exponential function is part of the integral.