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Magazine Chapter 6: Problem 11
Use integration by parts to find each integral. $$ \int x^{5} \ln x d x $$
Short Answer
Expert verified
The integral is \( \frac{x^6}{6} \ln x - \frac{x^6}{36} + C \).
Step by step solution
01
Choose Parts for Integration by Parts
Identify functions for integration by parts. Let \( u = \ln x \) and \( dv = x^5 \, dx \). This choice is because when differentiating \( \ln x \), it simplifies to \( \frac{1}{x} \), making it easier to integrate than differentiating \( x^5 \).
02
Differentiate and Integrate Parts
Differentiate \( u \) to find \( du \): \( du = \frac{1}{x} \, dx \). Integrate \( dv \) to find \( v \): \( v = \frac{x^6}{6} \).
03
Apply Integration by Parts Formula
Use the integration by parts formula \( \int u \, dv = uv - \int v \, du \). Substitute \( u = \ln x \), \( du = \frac{1}{x} \, dx \), \( v = \frac{x^6}{6} \), and \( dv = x^5 \, dx \).
04
Substitute and Simplify
Substitute into the formula: \( \int x^5 \ln x \, dx = \left( \ln x \cdot \frac{x^6}{6} \right) - \int \left( \frac{x^6}{6} \cdot \frac{1}{x} \right) \, dx \). Simplify to \( \frac{x^6 \ln x}{6} - \frac{1}{6} \int x^5 \, dx \).
05
Integrate Remaining Integral
Solve \( \int x^5 \, dx \) which is \( \frac{x^6}{6} + C_1 \). Substitute back: \( \int x^5 \ln x \, dx = \frac{x^6 \ln x}{6} - \frac{x^6}{36} + C \).
06
Write Final Solution
Combine and simplify: \( \int x^5 \ln x \, dx = \frac{x^6}{6} \ln x - \frac{x^6}{36} + C \), where \( C \) is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
When you think about definite integrals, imagine calculating the exact area under a curve on a graph between two specific points. This area doesn't change; it's fixed. A definite integral is expressed with limits of integration, typically written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the boundaries of the interval.
Definite integrals are quite powerful because they allow us to calculate numerous real-world phenomena, such as:
Definite integrals are quite powerful because they allow us to calculate numerous real-world phenomena, such as:
- The distance an object travels over a certain period.
- The total accumulated quantity, like water flow over time.
- You start by finding the antiderivative or the indefinite integral of the function.
- Then, apply the Fundamental Theorem of Calculus, which states you must substitute the bounds into this antiderivative and subtract the results: \( F(b) - F(a) \).
Indefinite Integrals
Indefinite integrals might seem like they have a fancy name, but they're really about finding the antiderivative of a function. When we speak of an indefinite integral, we mean there's no set interval or bounds; it's like computing the general form of the area under a curve. Instead of a fixed number, the result includes a constant \( C \), written as \( \int f(x) \, dx = F(x) + C \).
This constant \( C \) is very important:
This constant \( C \) is very important:
- It represents any constant that could be added since differentiating a constant gives zero.
- This means indefinite integrals provide a family of solutions rather than a single answer.
- Identify the function you need to integrate.
- Use integration techniques, such as substitution or integration by parts, to find the antiderivative.
- Don't forget to add the constant \( C \) at the end of your result.
Techniques of Integration
There are several techniques to solve integrals, especially when you encounter more complex functions. Mastering these techniques will help you tackle a variety of integration problems. Let's go through some common ones:
Integration by Parts, as you saw in the exercise, allowed us to tackle the integral \( \int x^5 \ln x \, dx \) by carefully choosing \( u = \ln x \) and \( dv = x^5 \, dx \), simplifying a complex problem into manageable parts.
- Integration by Parts: This is like the product rule for derivatives. It is useful when integrating products of functions. The formula is \( \int u \, dv = uv - \int v \, du \). Choose \( u \) and \( dv \) such that \( du \) and \( v \) are simpler to work with.
- Substitution: This method is similar to using the chain rule in differentiation. Use substitution when you see a function inside another function (composite functions). We introduce a new variable to simplify the integral.
- Partial Fractions: Use this technique for rational functions, which are quotients of polynomials. This involves breaking down a complex rational function into simpler fractions that are easier to integrate.
Integration by Parts, as you saw in the exercise, allowed us to tackle the integral \( \int x^5 \ln x \, dx \) by carefully choosing \( u = \ln x \) and \( dv = x^5 \, dx \), simplifying a complex problem into manageable parts.
