/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Calculate the integrals.$$\int \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 39

Calculate the integrals.$$\int \frac{1}{(x+2)(x+3)} d x$$.

Short Answer

Expert verified
The integral is \( \ln\left|\frac{x+2}{x+3}\right| + C \).

Step by step solution

01

Set Up Partial Fraction Decomposition

To calculate the integral \( \int \frac{1}{(x+2)(x+3)} \, dx \), begin by expressing the integrand as a sum of partial fractions. Assume \( \frac{1}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3} \).
02

Solve for Coefficients A and B

Multiply both sides of the equation by \((x+2)(x+3)\) to eliminate the denominators:\[ 1 = A(x+3) + B(x+2) \] Now, expand and collect like terms:\[ 1 = (A+B)x + (3A + 2B) \] Compare the coefficients with the left side (which is 1 with no \(x\) terms):- Constant term: \(3A + 2B = 1\)- Coefficients of \(x\): \(A + B = 0\)
03

Solve the System of Equations

From the equations:1. \(A + B = 0\)2. \(3A + 2B = 1\)You can express \(B\) from the first equation: \(B = -A\). Substitute into the second equation:\[ 3A + 2(-A) = 1 \]\[ 3A - 2A = 1 \]\[ A = 1 \]Substitute \(A = 1\) back into \(B = -A\):\[ B = -1 \]
04

Rewrite the Integrand Using Partial Fractions

Substitute \(A = 1\) and \(B = -1\) into the partial fractions:\[ \frac{1}{(x+2)(x+3)} = \frac{1}{x+2} - \frac{1}{x+3} \]
05

Integrate Each Term Separately

Now integrate the expression:\[ \int \left( \frac{1}{x+2} - \frac{1}{x+3} \right) \, dx = \int \frac{1}{x+2} \, dx - \int \frac{1}{x+3} \, dx \]The integrals of these terms are:\[ \ln|x+2| - \ln|x+3| + C \]
06

Simplify the Resulting Expression

Combine the logarithmic expressions into a single logarithm:\[ \ln\left|\frac{x+2}{x+3}\right| + C \]

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Fraction Decomposition
Partial fraction decomposition is a technique used in calculus to simplify complex rational expressions. It involves breaking down complex fractions into simpler fractions that are easier to integrate. In the exercise, we have the fraction \( \frac{1}{(x+2)(x+3)} \), which is decomposed into simpler parts.
  • The principle begins with assuming that the complex fraction can be expressed as a sum of simpler fractions. For example, we assume \( \frac{1}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3} \).
  • Next, we eliminate the denominators by multiplying through by the original denominator, \((x+2)(x+3)\), to change the equation into \( 1 = A(x+3) + B(x+2) \). This equation is then expanded and simplified to handle each coefficient separately.
  • By comparing coefficients, we set up and solve a system of equations to find the values of \( A \) and \( B \). Solving \( A + B = 0 \) and \( 3A + 2B = 1 \) yields \( A = 1 \) and \( B = -1 \).
Using these decomposed fractions simplifies the integration process, turning a complicated problem into simpler ones which are more straightforward to handle.
Definite and Indefinite Integrals
Integral calculus involves finding the integral of a function, which can be definite or indefinite. In this exercise, we evaluate the indefinite integral of a rational function.
  • An indefinite integral, denoted by \( \int \, dx \), represents a family of functions and includes a constant \( C \), representing the constant of integration. It is used when no specific interval is given.
  • A definite integral, on the other hand, calculates the net area under a curve within a specific range or interval \([a, b]\). It does not include a constant \( C \), as it results in a numerical value representing the area.
In the problem, once we transform \( \frac{1}{(x+2)(x+3)} \) into partial fractions, we handle the indefinite integral separately for each fraction. The integration process here doesn't involve limits, making it indefinite. Integrating \( \int \frac{1}{x+2} \, dx - \int \frac{1}{x+3} \, dx \) gives us the solution with a logarithmic component and the constant \( C \). This process is essential in integral calculus for solving functions symbolically rather than numerically.
Logarithmic Integration
In integral calculus, logarithmic integration refers to the process of integrating functions that result in natural logarithms. This is particularly relevant for rational functions where the integrand is of the form \( \frac{1}{x+a} \).
  • When integrating functions such as \( \frac{1}{x+a} \), the result is the natural logarithm \( \ln|x+a| \). In our exercise, the integration of \( \frac{1}{x+2} \) results in \( \ln|x+2| \), while \( \frac{1}{x+3} \) yields \( \ln|x+3| \).
  • Logarithmic integration helps to simplify and solve otherwise complex rational functions. It makes it easier to see the connection between the derivative and integral operations.
After deriving \( \ln|x+2| - \ln|x+3| + C \) from the individual integrals, we simplify using properties of logarithms. Specifically, \( \ln a - \ln b = \ln(\frac{a}{b}) \), giving us a final expression of \( \ln\left|\frac{x+2}{x+3}\right| + C \). This simplification demonstrates how logarithmic identities play an essential role in manipulating and understanding integral results.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find \(\int 4 x\left(x^{2}+1\right) d x\) using two methods: (a) Do the multiplication first, and then antidifferentiate. (b) Use the substitution \(w=x^{2}+1\) (c) Explain how the expressions from parts (a) and (b) are different. Are they both correct?

Throughout much of the \(20^{\text {th }}\) century, the yearly consumption of electricity in the US increased exponentially at a continuous rate of \(7 \%\) per year. Assume this trend continues and that the electrical energy consumed in 1900 was 1.4 million megawatt-hours. (a) Write an expression for yearly electricity consumption as a function of time, \(t,\) in years since \(1900 .\) (b) Find the average yearly electrical consumption throughout the \(20^{\text {th }}\) century. (c) During what year was electrical consumption closest to the average for the century? (d) Without doing the calculation for part (c), how could you have predicted which half of the century the answer would be in?

Explain why the two antiderivatives are really, despite their apparent dissimilarity, different expressions of the same problem. You do not need to evaluate the integrals. $$\int \frac{\ln x}{x} d x \text { and } \, \int x \, d x$$

Find the exact area of the regions.Bounded by \(y=x^{2} / \sqrt{1-x^{2}}, y=0, x=0, x=1 / 2\).

Decide whether the statements are true or false. Give an explanation for your answer.To calculate \(\int \frac{1}{x^{3}+x^{2}} d x,\) we can split the integrand into $$\int\left(\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}\right) d x$$.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.