/*! 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 76 Solve without using integral tab... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 76

Solve without using integral tables. Calculate the integral \(\int \frac{1}{(x-a)(x-b)} d x\) for (a) \(a \neq b\) (b) \(a=b\)

Short Answer

Expert verified
For (a) \( \int \frac{1}{(x-a)(x-b)} \, dx = \frac{1}{a-b} \ln \left| \frac{x-a}{x-b} \right| + C \). For (b), \( \int \frac{1}{(x-a)^2} \, dx = -\frac{1}{x-a} + C \).

Step by step solution

01

Recognize the Integral Type and Initial Approach for (a)

The integral \( \int \frac{1}{(x-a)(x-b)} \, dx \) suggests the use of partial fraction decomposition because its denominator is a product of linear factors. Since \( a eq b \), the expression can be split into simpler fractions.
02

Set Up Partial Fraction Decomposition for (a)

Express the integrand as the sum of partial fractions: \( \frac{1}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b} \). Multiply through by the denominator \((x-a)(x-b)\) to get \(1 = A(x-b) + B(x-a)\).
03

Solve for Coefficients A and B (a)

To find \(A\) and \(B\), substitute convenient values for \(x\). First, set \(x = a\) to eliminate \(B\): \(1 = A(a-b) \Rightarrow A = \frac{1}{a-b}\).Then, set \(x = b\) to eliminate \(A\):\(1 = B(b-a) \Rightarrow B = \frac{-1}{a-b}\).
04

Integrate Each Partial Fraction (a)

Substitute \(A\) and \(B\) back into the partial fractions:\(\int \frac{1}{(x-a)(x-b)} \, dx = \int \left( \frac{1}{a-b} \cdot \frac{1}{x-a} - \frac{1}{a-b} \cdot \frac{1}{x-b} \right) \, dx\).This splits into two easy integrals:\(= \frac{1}{a-b} \ln |x-a| - \frac{1}{a-b} \ln |x-b| + C\).
05

Simplify the Result (a)

Combine the logarithms:\( \frac{1}{a-b}(\ln|x-a| - \ln|x-b|) = \frac{1}{a-b} \ln \left| \frac{x-a}{x-b} \right| + C\).
06

Analyze Case for (b), when \(a = b\)

When \(a = b\), the expression \((x-a)(x-b) = (x-a)^2\) and the integral becomes \(\int \frac{1}{(x-a)^2} \, dx\). This form requires a different approach.
07

Integrate for Case (b)

The integral \(\int \frac{1}{(x-a)^2} \, dx\) is a standard form. The antiderivative of \(\frac{1}{u^2}\) where \(u = x-a\) is \(-\frac{1}{u} + C\). Thus, \(\int \frac{1}{(x-a)^2} \, dx = -\frac{1}{x-a} + C\).

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 to break down complex rational expressions into simpler ones, making integration easier. This tool is particularly useful when dealing with integrals where the denominator is a product of linear or quadratic terms. By decomposing such expressions, they can be rewritten as a sum of fractions that are easier to integrate.

Here's how it works:
  • Identify the type of fraction: Ensure your denominator is a product of distinct linear factors, as seen in \(\frac{1}{(x-a)(x-b)}\).
  • Set up the decomposition: Express the fraction as a sum of individual fractions, like \(\frac{1}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}\).
  • Clear the denominator: Multiply through by the common denominator to eliminate fractions. This represents the equation \(1 = A(x-b) + B(x-a)\).
  • Solve for coefficients: Use strategic values of \(x\) to solve for \(A\) and \(B\), often setting \(x = a\) and \(x = b\) to simplify calculation.
This breakdown makes it straightforward to integrate each component separately, as the partial fractions are more manageable.
Integration Techniques
When integrating functions, employing the right integration technique is crucial for simplifying calculations. In this context, the partial fraction decomposition is an integration technique that transforms a complex integrand into simpler components.

Once the decomposition is achieved:
  • Integrate each term individually: Each term in the partial fraction has a standard integration form. For instance, \( \int \frac{1}{x-a} \, dx = \ln |x-a| + C\).
  • Combine results: After integrating individual fractions, combine the results. For our example, the integration results in terms like \(\frac{1}{a-b} \ln|x-a| - \frac{1}{a-b} \ln|x-b|\).
  • Simplify: Simplify expressions further using logarithmic identities when possible, resulting in cleaner, more elegant solutions like \(\frac{1}{a-b} \ln \left| \frac{x-a}{x-b} \right| + C\).
Combining partial fraction decomposition with straightforward integration rules provides an effective strategy when tackling rational integrals.
Definite vs Indefinite Integrals
Understanding the difference between definite and indefinite integrals is essential in calculus.
  • Indefinite Integrals: These represent a family of functions, characterized by the integration constant \(C\), indicating an antiderivative of a function. The solution format like \(\ln \left| \frac{x-a}{x-b} \right| + C\) is an indefinite integral because it incorporates arbitrary constants.
  • Definite Integrals: These calculate the exact area under a curve between two limits, providing a single numerical value. Unlike indefinite integrals, definite integrals do not include \(C\) since the arithmetic of the limits cancels it out.
In tackling integrals, it's crucial to discern the type required by the problem to apply the correct method, whether evaluating specific boundaries or determining general antiderivatives.

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

Give an example of: An indefinite integral involving a square root that can be evaluated by first completing a square.

The rate at which water is flowing into a tank is \(r(t)\) gallons/minute, with \(t\) in minutes. (a) Write an expression approximating the amount of water entering the tank during the interval from time \(t\) to time \(t+\Delta t,\) where \(\Delta t\) is small. (b) Write a Riemann sum approximating the total amount of water entering the tank between \(t=0\) and \(t=5 .\) Write an exact expression for this amount. (c) By how much has the amount of water in the tank changed between \(t=0\) and \(t=5\) if \(r(t)=20 e^{a+2 t} ?\) (d) If \(r(t)\) is as in part (c), and if the tank contains 3000 gallons initially, find a formula for \(Q(t),\) the amount of water in the tank at time \(t\)

Evaluate the indefinite integral, using a trigonometric substitution and a triangle to express the answer in terms of \(x\).$$\int \frac{x^{3}}{\sqrt{4-x^{2}}} d x$$.

Find the exact area. Under \(f(x)=1 /(x+1)\) between \(x=0\) and \(x=2\)

Using the substitution \(w=e^{x},\) find a function \(g(w)\) such that \(\int_{a}^{b} e^{-x} d x=\int_{e^{a}}^{e^{b}} g(w) d w\) for all \(a
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Pure Maths

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.