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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.2.38

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫ x⁵ e³ˣ dx

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Identify the integral to solve: \(\int x^{5} e^{3x} \, dx\).
Recognize that this integral involves a product of a polynomial \(x^{5}\) and an exponential function \(e^{3x}\), which suggests using integration by parts.
Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). Choose \(u = x^{5}\) (since its derivative simplifies) and \(dv = e^{3x} \, dx\) (since it integrates easily).
Compute \(du\) and \(v\): differentiate \(u\) to get \(du = 5x^{4} \, dx\), and integrate \(dv\) to get \(v = \frac{1}{3} e^{3x}\).
Apply the integration by parts formula: \(\int x^{5} e^{3x} \, dx = x^{5} \cdot \frac{1}{3} e^{3x} - \int \frac{1}{3} e^{3x} \cdot 5x^{4} \, dx\). This reduces the original integral to a similar integral with a lower power of \(x\), which you can solve by repeating integration by parts until the polynomial term is eliminated.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation. It transforms the integral of a product of functions into simpler integrals, using the formula ∫u dv = uv - ∫v du. This method is especially useful when integrating products like x⁵ and e³ˣ.
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Choosing u and dv

Selecting which part of the integrand to assign as u and which as dv is crucial for simplifying the integral. Typically, u is chosen as a polynomial (like x⁵) because its derivative simplifies, and dv as an exponential or trigonometric function, which is easy to integrate.
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Repeated Application of Integration by Parts

When the polynomial degree is high, integration by parts may need to be applied multiple times, reducing the power of x step-by-step. This iterative process continues until the polynomial term is eliminated, allowing the integral to be evaluated fully.
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Repeated Integration by Parts