Chapter 29: Problem 8
Determine the following: $$ \int \frac{x^{2}}{x+1} d x $$
Short Answer
Expert verified
The short answer to the given integral is:
\( \int \frac{x^{2}}{x+1} d x = \frac{1}{2}(x+1)^2 - 2(x+1) + \ln |x+1| + C \)
Step by step solution
01
Choose the substitution variable and differential
Let's define a new variable, \(u\), as
$$
u = x + 1
$$
Now, differentiate both sides with respect to \(x\). We get
$$
\frac{du}{dx} = 1
$$
Which implies that
$$
d u = d x
$$
02
Rewrite the integration in terms of \(u\)
Since \(u = x + 1\), we can rewrite \(x^2\) as
$$
x^{2} = (u - 1)^{2}
$$
Now, substitute this expression into the integral
$$
\int \frac{x^{2}}{x+1} d x = \int \frac{(u-1)^2}{u} du
$$
03
Perform the integration
Now integrate the new expression in terms of \(u\)
$$
\int \frac{(u-1)^2}{u} du = \int \frac{u^2-2u+1}{u} du
$$
Now we can split the integral into separate fractions:
$$
\int \frac{u^2-2u+1}{u} du = \int u du - 2\int du + \int \frac{1}{u} du
$$
Integrate each part:
$$
\int u du = \frac{1}{2}u^2 \\
\int du = u \\
\int \frac{1}{u} du = \ln |u|
$$
Put the results together:
$$
\frac{1}{2}u^2 - 2u + \ln |u| + C = \frac{1}{2}(x+1)^2 - 2(x+1) + \ln |x+1| + C
$$
The constant \(C\) represents the constant of integration. Our solution to the given integral is:
$$
\int \frac{x^{2}}{x+1} d x = \frac{1}{2}(x+1)^2 - 2(x+1) + \ln |x+1| + C
$$
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.
Definite Integrals
Definite integrals are a fundamental concept in calculus where you calculate the net area under a curve between two specific points. Unlike indefinite integrals, which represent a family of functions, definite integrals provide a specific numerical value.
- This value represents the accumulation of quantities, such as total distance over time or total area under a curve.
- Definite integrals require limits of integration, typically denoted as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the points of integration.
Indefinite Integrals
Indefinite integrals are a key player in calculus and are used to find the antiderivative of a function. This operation is essentially the reverse of differentiation, often referred to as integration. While solving an indefinite integral, you generally include a constant of integration, \(C\), because integration is not unique.
- The indefinite integral does not have limits of integration—it describes a general form of antiderivatives.
- The process usually involves reversing differentiation rules to find a function that, when derived, yields the original function.
Change of Variables Technique
The change of variables technique, also known as substitution, is a powerful method used to simplify integration. It is particularly useful when dealing with integrals that are otherwise difficult or impossible to evaluate directly.The central idea is to introduce a new variable, commonly denoted as \(u\), which replaces a more complex function of \(x\) in the integral. This substitution often simplifies the function, making it easier to integrate.
- Start by choosing a substitution, such as \(u = g(x)\), where \(g(x)\) is part of the integrand.
- Differentiate \(u\) with respect to \(x\), forming \(\frac{du}{dx}\) to express \(dx\) in terms of \(du\).
- Rewrite the integral with \(u\), replacing \(x\) and \(dx\), which often results in a simpler form to integrate.