/*! 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 23 Find each integral by using the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 23

Find each integral by using the integral table on the inside back cover. $$ \int \frac{1}{x(x+3)} d x $$

Short Answer

Expert verified
\( \frac{1}{3} \ln|x| - \frac{1}{3} \ln|x+3| + C \)

Step by step solution

01

Decompose the function

We need to express \( \frac{1}{x(x+3)} \) as a sum of partial fractions to integrate it easily. Assume \( \frac{1}{x(x+3)} = \frac{A}{x} + \frac{B}{x+3} \). To find \( A \) and \( B \), multiply through by the denominator to get \( 1 = A(x+3) + Bx \).
02

Solve for the coefficients

Expanding and collecting like terms gives \( 1 = Ax + 3A + Bx \). Gather like terms: \( (A + B)x + 3A = 1 \). Therefore, we equate coefficients: \( A+B = 0 \) and \( 3A = 1 \). Solve to find: \( A = \frac{1}{3} \) and \( B = -\frac{1}{3} \).
03

Substitute back into the integral

Substitute \( A \) and \( B \) back into the partial fractions: \( \frac{1}{x(x+3)} = \frac{1/3}{x} - \frac{1/3}{x+3} \). This decomposes the integral: \[\int \frac{1}{x(x+3)} \, dx = \int \frac{1/3}{x} \, dx - \int \frac{1/3}{x+3} \, dx\]
04

Integrate each term separately

Use the fact that the integral of \( \frac{1}{x} \) is \( \ln|x| \). Therefore: \[\int \frac{1}{3x} \, dx = \frac{1}{3} \ln|x|\] and \[\int \frac{1}{3(x+3)} \, dx = \frac{1}{3} \ln|x+3|\]
05

Combine results

Subtract the results from each integral: \[\frac{1}{3} \ln|x| - \frac{1}{3} \ln|x+3| + C\] where \( C \) 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 91影视!

Key Concepts

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

Partial Fraction Decomposition
When faced with complex rational functions, Partial Fraction Decomposition becomes a valuable tool. It allows us to break down a fraction into simpler parts, making integration easier. Consider a function like \( \frac{1}{x(x+3)} \). By expressing it as a sum of two fractions, \( \frac{A}{x} + \frac{B}{x+3} \), we simplify the expression into components that are more straightforward to manage.
To apply partial fraction decomposition:
  • Express the complex fraction as a sum of simpler fractions.
  • Assume denominators corresponding to each factor of the original denominator (e.g., \(x\) and \(x+3\)).
  • Set up an equation where the original fraction equals the sum of the decomposed fractions.
  • Multiply through by the denominator to clear fractions and equate coefficients to solve for unknowns \(A\) and \(B\).
This step-by-step breakdown allows for the integration of each term separately, paving the way for easier calculations.
Integral Calculus
Integral Calculus helps us find the accumulated area under curves or solve differential problems. It plays a vital role in calculating the total change from a series of infinitesimally small increments. In our problem example, we find the integral of \( \frac{1}{x(x+3)} \) by breaking it into simpler fractions using partial fraction strategies.
This involves integration techniques such as:
  • Algebraically manipulating functions into simpler components.
  • Recognizing standard integral forms to simplify the process.
  • Applying known results of elementary integrals like \( \ln|x| \) for \( \frac{1}{x} \).
Integral calculus hence supports tackling advanced computations by breaking them down into standard integral forms, ultimately simplifying the whole integration process.
Indefinite Integrals
Indefinite integrals represent the antiderivative of a function and result in a family of functions with an added constant of integration \(C\). They lie at the heart of many calculus solutions, providing general solutions to functions' differential equations.
To compute these integrals, particularly when expressed as partial fractions:
  • Separate the problem into manageable parts that correspond to recognizable integrals.
  • Use integral rules for straightforward integrations, like converting a function into \( \ln|x| \) when integrating \( \frac{1}{x} \).
  • Add the constant \(C\) to represent the family of antiderivatives.
Undoubtedly, indefinite integrals offer foundational insight to formulating solutions across diverse mathematical problems, allowing for vast applications in mathematical modeling and beyond.

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

After reading the preceding Example, find each integral by repeated integration by parts using a table. $$ \int(x+1)^{2}(x+2)^{5} d x $$

Suppose that you meet 30 new people each year, but each year you forget \(20 \%\) of all of the people that you know. If \(y(t)\) is the total number of people who you remember after \(t\) years, then \(y\) satisfies the differential equation \(y^{\prime}=30-0.2 y .\) (Do you see why?) Solve this differential equation subject to the condition \(y(0)=0 \quad\) (you knew no one at birth).

The following are differential equations stated in words. Find the general solution of each. The derivative of a function at each point is \(-2\).

$$ \text { Use integration by parts to find each integral. } $$ $$ \int t e^{-0.2 t} d t $$

In the reservoir model, the heart is viewed as a balloon that swells as it fills with blood (during a period called the systole), and then at time \(t_{0}\) it shuts a valve and contracts to force the blood out (the diastole). Let \(p(t)\) represent the pressure in the heart at time \(t\) a. During the diastole, which lasts from \(t_{0}\) to time \(T, p(t)\) satisfies the differential equation $$\frac{d p}{d t}=-\frac{K}{R} p$$ Find the general solution \(p(t)\) of this differential equation. ( \(K\) and \(R\) are positive constants determined, respectively, by the strength of the heart and the resistance of the arteries. The differential equation states that as the heart contracts, the pressure decreases \((d p / d t\) is negative) in proportion to itself.) b. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0} . \quad\left(p_{0}\right.\) is a constant representing the pressure at the transition time \(t_{0}\).) c. During the systole, which lasts from time 0 to time \(t_{0}\), the pressure \(p(t)\) satisfies the differential equation $$\frac{d p}{d t}=K I_{0}-\frac{K}{R} p$$ Find the general solution of this differential equation. \(\left(I_{0}\right.\) is a positive constant representing the constant rate of blood flow into the heart while it is expanding.) [Hint: Use the same \(u\) -substitution technique that was used in Example 7.] d. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0}\). e. In parts (b) and (d) you found the formulas for the pressure \(p(t)\) during the diastole \(\left(t_{0} \leq t \leq T\right)\) and the systole \(\left(0 \leq t \leq t_{0}\right)\). Since the heart behaves in a cyclic fashion, these functions must satisfy \(p(T)=p(0)\). Equate the solutions at these times (use the correct formula for each time) to derive the important relationship $$ R=\frac{p_{0}}{I_{0}} \frac{1-e^{-K T / R}}{1-e^{-K t_{0} / R}} $$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Applied 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.