Chapter 9: Problem 16
Evaluate the following integrals: \(\int x^{5} \ln x d x\)
Short Answer
Expert verified
\(\int x^{5} \ln x dx = \frac{x^{6}}{6} \ln x - \frac{x^{6}}{36} + C\)
Step by step solution
01
Identify the methods
The given integral is \(\[\begin{equation} \int x^{5} \ln x d x \end{equation}\]\). This integral requires integration by parts, which is based on the formula \(\[\begin{equation} \int u \ dv = uv - \int v \ du \end{equation}\]\).
02
Choose u and dv
Let \(u = \ln x\) and \(dv = x^{5} dx\). Then, differentiate \(u\) and integrate \(dv\):\(du = \frac{1}{x} dx, \quad v = \int x^{5} dx = \frac{x^{6}}{6}\)\.
03
Substitute into the integration by parts formula
Substitute \(u, dv, du, \text{and} v\) into the integration by parts formula: \(\[\begin{equation} \int x^{5} \ln x \ dx = \frac{x^{6}}{6} \ln x - \int \frac{x^{6}}{6} \frac{1}{x} \ dx \end{equation}\]\).
04
Simplify the integral
Simplify the expression inside the remaining integral: \(\[\begin{equation} \int \frac{x^{6}}{6} \frac{1}{x} \ dx = \int \frac{x^{5}}{6} \ dx = \frac{1}{6} \int x^{5} \ dx \end{equation}\]\).\ Integrating \(x^{5}\): \(\[\begin{equation} \frac{1}{6} \int x^{5} \ dx = \frac{1}{6} \frac{x^{6}}{6} = \frac{x^{6}}{36}\end{equation}\]\).
05
Write the final answer
Combine the results from the integration by parts steps: \(\[\begin{equation} \int x^{5} \ln x \ dx = \frac{x^{6}}{6} \ln x - \frac{x^{6}}{36} + C \end{equation}\]\), where \(C \text{is the constant of integration} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
integration by parts
Integration by parts is a powerful technique used to integrate products of functions. It is based on the product rule of differentiation. The formula is: \[ \int u \ dv = uv - \int v \ du \]Here are the steps to use integration by parts:
- Choose which part of the integrand to call \(u\) and which to call \(dv\).
- Differentiatiate \(u\) to get \(du\), and integrate \(dv\) to get \(v\).
- Substitute these into the formula \(uv - \int v \ du\).
- Integrate the remaining integral.
definite and indefinite integrals
Integrals are classified as definite or indefinite:
- Definite integrals have specific limits of integration and represent the signed area under the curve of a function between two points.
- Indefinite integrals do not have limits and represent a family of functions, including an arbitrary constant \(C\).
logarithmic functions
Logarithmic functions are the inverse of exponential functions. They have the form \( \log_b(x) \) and are commonly used in many mathematical applications.
- Natural logarithms (\( \ln x \)) use the base \(e\), where \(e \approx 2.718\).
differentiation and integration
Differentiation and integration are fundamental concepts in calculus, acting as inverse processes.
- Differentiation finds the rate at which a function is changing at any point.
- Integration finds the total accumulation of quantities, such as area under a curve.