/*! 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 15 Evaluate the integral. \(\int \f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Evaluate the integral. \(\int \frac{x-1}{x^{2}+2 x} d x\)

Short Answer

Expert verified
The integral evaluates to \(-\frac{1}{2} \ln |x| + \frac{3}{2} \ln |x+2| + C\).

Step by step solution

01

Identify the Method of Integration

To evaluate \(\int \frac{x-1}{x^2+2x} \, dx\), recognize that the integrand can be decomposed using partial fraction decomposition since the denominator is factorable.
02

Factor the Denominator

The denominator \(x^2 + 2x\) can be factored as \(x(x+2)\). Thus, we rewrite the integrand with this denominator in mind.
03

Set Up Partial Fraction Decomposition

Express \(\frac{x-1}{x(x+2)}\) as \(\frac{A}{x} + \frac{B}{x+2}\). We need to find the values of \(A\) and \(B\).
04

Solve for A and B

Multiply through by the common denominator to clear the fractions: \(x-1 = A(x+2) + Bx\). Expand and equate coefficients: \(x - 1 = Ax + 2A + Bx\). Thus, \(A + B = 1\) and \(2A = -1\). Solve these equations to get \(A = -\frac{1}{2}\) and \(B = \frac{3}{2}\).
05

Rewrite the Integral

Substitute the values of \(A\) and \(B\) back into the partial fractions: \(\int \left( \frac{-1/2}{x} + \frac{3/2}{x+2} \right) \, dx\).
06

Integrate Each Term Separately

Integrate the terms separately: \(-\frac{1}{2} \int \frac{1}{x} \, dx + \frac{3}{2} \int \frac{1}{x+2} \, dx\). The antiderivatives are: \(-\frac{1}{2} \ln |x| + \frac{3}{2} \ln |x+2| + C\), where \(C\) is the integration constant.
07

Combine the Logarithms

Use the properties of logarithms to combine the two log terms: \(\ln \left( \frac{|x+2|^{3/2}}{|x|^{1/2}} \right) + C\). This is the final simplified result.

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.

Integration Techniques
Integration is a mathematical technique that allows us to find the area under curves or the total accumulation of quantities. Different functions require different approaches to integration. In this exercise, we use **Partial Fraction Decomposition**, a technique often employed when dealing with rational functions. A rational function is one where the numerator and the denominator are both polynomials. The method involves expressing a complex rational function as a sum of simpler fractions, making it easier to integrate each component separately.
This technique becomes especially useful when the denominator of a rational function is factorable, as seen in the example with the integral \( \int \frac{x-1}{x^2+2x} dx \). By factoring the denominator and expressing the integrand using partial fractions, we can transform the integral into a more manageable form.
Logarithmic Integration
Logarithmic integration refers to the integration of terms involving natural logarithms. In our example exercise, the expressions like \( \frac{1}{x} \) and \( \frac{1}{x+2} \) lend themselves to logarithmic integration.
When integrating \( \frac{1}{x} \) or similar terms \( \frac{1}{ax+b} \), the result is a natural logarithm. Specifically, \( \int \frac{1}{x} dx = \ln|x| + C \). Similarly, \( \int \frac{1}{x+2} dx = \ln|x+2| + C \).
These results come from the fundamental property of logarithms in calculus: differentiating a logarithmic function \( \ln u \) yields \( \frac{1}{u} \), making integration the inverse operation.
Factorization
Factorization is the process of breaking down an expression into a product of simpler expressions or factors. It is a crucial step in many integration problems, particularly when working with rational functions. In our problem, the expression \( x^2 + 2x \) is factorable.
We can write it as \( x \cdot (x+2) \). Recognizing this allows us to use **Partial Fraction Decomposition**, simplifying the integration process.
When the denominator of a rational function is factorable, we can decompose the rational expression into simpler fractions, ultimately simplifying the integration into basic logarithmic forms.
Integration Constants
Integration often introduces an arbitrary constant known as the integration constant, denoted as \( C \). This constant is crucial because it accounts for the fact that functions that differ by a constant have the same derivative.
In the final step of the integration process, such as finding the antiderivative of a function, the integration constant is included. For example, the indefinite integral \( \int \frac{1}{x} dx \) results in \( \ln|x| + C \).
The constant \( C \) represents the infinite set of parallel curves that could satisfy the given differential equation, allowing the integration to encompass all possible solutions.

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

Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson's Rule to approximate the given integral with the specified value of \( n \). (Round your answers to six decimal places.) \( \displaystyle \int_0^4 \ln (1 + e^x)\ dx \) , \( n = 8 \)

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \( \displaystyle \int_0^5 \frac{1}{\sqrt[3]{5 - x}}\ dx \)

Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson's Rule to approximate the given integral with the specified value of \( n \). (Round your answers to six decimal places.) \( \displaystyle \int_0^4 \sqrt{y} \cos y\ dy \) , \( n = 8 \)

The integral $$ \int_0^\infty \frac{1}{\sqrt{x} (1 + x)}\ dx $$ is improper for two reasons: The interval \( [0, \infty) \) is infinite and the integrand has an infinite discontinuity at 0. Evaluate it by expressing it as a sum of improper integrals of Type 2 and Type 1 as follows: $$ \int_0^\infty \frac{1}{\sqrt{x} (1 + x)}\ dx = \int_0^1 \frac{1}{\sqrt{x} (1 + x)}\ dx + \int_1^\infty \frac{1}{\sqrt{x} (1 + x)}\ dx $$

Find the approximations \( L_n \), \( R_n \), \( T_n \), and \( M_n \) for \( n = 5 \), 10, and 20. Then compute the corresponding errors \( E_L \), \( E_R \), \( E_T \), and \( E_M \). (Round your answers to six decimal places. You may wish to use the sum command on a computer algebra system.) What observations can you make? In particular, what happens to the errors when \( n \) is doubled? \( \displaystyle \int_1^2 \frac{1}{x^2}\ dx \)

See all solutions

Recommended explanations on Math Textbooks

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.