/*! 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 21 Evaluate the following integrals... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 21

Evaluate the following integrals. $$\int \frac{6 x^{2}}{x^{4}-5 x^{2}+4} d x$$

Short Answer

Expert verified
Answer: The integral is \(-6 \ln |x + 2| + 6 \ln |x - 2| + 2 \ln |x + 1| + 2 \ln |x - 1| + C\), where \(C\) is the integration constant.

Step by step solution

01

Factor the denominator

First, let's factor the denominator of the integrand: $$x^4 - 5x^2 + 4 = (x^2 - 4)(x^2 - 1) = (x + 2)(x - 2)(x + 1)(x - 1)$$
02

Decompose into partial fractions

Now we decompose the integrand into partial fractions: $$\frac{6x^2}{(x + 2)(x - 2)(x + 1)(x - 1)} = \frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C}{x + 1} + \frac{D}{x - 1}$$ To find the constants \(A, B, C\), and \(D\), we will clear the fractions and then substitute suitable values for \(x\) to make unknowns zero.
03

Clear fractions and find constants

First, clear fractions: $$6x^2 = A(x - 2)(x + 1)(x - 1) + B(x + 2)(x + 1)(x - 1) + C(x + 2)(x - 2)(x - 1) + D(x + 2)(x - 2)(x + 1)$$ Now find the constants by substituting suitable values for \(x\): 1. If \(x = -2\), then \(6(-2)^2 = 24\), and we get \(24 = -4A \Rightarrow A = -6\) 2. If \(x = 2\), then \(6(2)^2 = 24\), and we get \(24 = 4B \Rightarrow B = 6\) 3. If \(x = -1\), then \(6(-1)^2 = 6\), and we get \(6 = 3C \Rightarrow C = 2\) 4. If \(x = 1\), then \(6(1)^2 = 6\), and we get \(6 = 3D \Rightarrow D = 2\) So we have: $$\frac{6x^2}{(x + 2)(x - 2)(x + 1)(x - 1)} = \frac{-6}{x + 2} + \frac{6}{x - 2} + \frac{2}{x + 1} + \frac{2}{x - 1}$$
04

Integrate each term

Now, we integrate each term separately: \begin{align*} \int \frac{6 x^{2}}{x^{4}-5 x^{2}+4} d x &= \int \left(\frac{-6}{x + 2} + \frac{6}{x - 2} + \frac{2}{x + 1} + \frac{2}{x - 1}\right) dx \\ &= -6 \int \frac{1}{x + 2} dx + 6 \int \frac{1}{x - 2} dx + 2 \int \frac{1}{x + 1} dx + 2 \int \frac{1}{x - 1} dx \\ &= -6 \ln |x + 2| + 6 \ln |x - 2| + 2 \ln |x + 1| + 2 \ln |x - 1| + C \\ \end{align*} Where \(C\) is the integration constant. So the final answer is: $$\int \frac{6 x^{2}}{x^{4}-5 x^{2}+4} d x = -6 \ln |x + 2| + 6 \ln |x - 2| + 2 \ln |x + 1| + 2 \ln |x - 1| + C$$

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

Let \(I_{n}=\int x^{n} e^{-x^{2}} d x,\) where \(n\) is a nonnegative integer. a. \(I_{0}=\int e^{-x^{2}} d x\) cannot be expressed in terms of elementary functions. Evaluate \(I_{1}\). b. Use integration by parts to evaluate \(I_{3}\). c. Use integration by parts and the result of part (b) to evaluate \(I_{5}\). d. Show that, in general, if \(n\) is odd, then \(I_{n}=-\frac{1}{2} e^{-x^{2}} p_{n-1}(x)\) where \(p_{n-1}\) is a polynomial of degree \(n-1\). e. Argue that if \(n\) is even, then \(I_{n}\) cannot be expressed in terms of elementary functions.

Use integration by parts to evaluate the following integrals. $$\int_{0}^{\infty} x e^{-x} d x$$

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

Find the volume of the solid torus formed when the circle of radius 4 centered at (0,6) is revolved about the \(x\) -axis.

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \sqrt{e^{x}+1} d x \text { (Hint: Let } u=\sqrt{e^{x}+1}$$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics 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.