/*! 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 29 Evaluate the following integrals... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 29

Evaluate the following integrals. $$\int \frac{x+2}{x+4} d x$$

Short Answer

Expert verified
Question: Evaluate the integral of the function $\frac{x+2}{x+4}$. Answer: The integral of the function $\frac{x+2}{x+4}$ is given by $\int \frac{x+2}{x+4} d x = x - 2 \ln|x+4| + C$, where \(C\) is the constant of integration.

Step by step solution

01

Division and Simplification

First, perform polynomial long division: $$\frac{x+2}{x+4} = 1-\frac{2}{x+4}.$$ Now, the given integral can be written as $$\int \frac{x+2}{x+4} d x = \int \left( 1-\frac{2}{x+4} \right) d x.$$
02

Integrating the simplified expression

Now integrate term by term: $$\int \left( 1-\frac{2}{x+4} \right) d x = \int 1 d x - \int \frac{2}{x+4} d x.$$
03

Separating the integration terms and finding antiderivatives

For the first part, $$\int 1 dx = x + C_1,$$ where \(C_1\) is a constant of integration. For the second part, $$\int \frac{2}{x+4} dx = 2 \int \frac{1}{x+4} dx.$$ Here, we can make a substitution using \(u = x + 4\) and calculating \(du/dx\): $$\frac{d u}{d x} = \frac{d (x+4)}{d x} = 1,$$ and therefore \(d u = d x\).
04

Substitution and reverse substitution

Substitute \(u\) and \(du\) in the second integral: $$2 \int \frac{1}{x+4} dx = 2 \int \frac{1}{u} du = 2 \ln|u| + C_2,$$ where \(C_2\) is another constant of integration. Now substitute back \(x + 4\) for \(u\): $$2 \ln|u| + C_2 = 2 \ln|x+4| + C_2.$$
05

Final answer

Combine both integrals and the constants of integration: $$\int \left( 1-\frac{2}{x+4} \right) d x = \left( x + C_1 \right) - \left( 2 \ln|x+4| + C_2 \right) = x - 2 \ln|x+4| + C,$$ where \(C = C_1 - C_2\) is a constant. Hence, $$\int \frac{x+2}{x+4} d x = x - 2 \ln|x+4| + C.$$

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!

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

The heights of U.S. men are normally distributed with a mean of 69 inches and a standard deviation of 3 inches. This means that the fraction of men with a height between \(a\) and \(b\) (with \(a

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \cos x \cos 2 x d x$$

The Mercator map projection was proposed by the Flemish geographer Gerardus Mercator \((1512-1594) .\) The stretching of the Mercator map as a function of the latitude \(\theta\) is given by the function $$ G(\theta)=\int_{0}^{\theta} \sec x d x $$ Graph \(G,\) for \(0 \leq \theta<\pi / 2\)

Refer to the summary box (Partial Fraction Decompositions) and evaluate the following integrals. $$\int \frac{x^{3}+1}{x\left(x^{2}+x+1\right)^{2}} d x$$

Evaluate the following integrals. Consider completing the square. $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-1)(x-3)}}$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Calculus

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.