/*! 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 49 Use the approaches discussed in ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 49

Use the approaches discussed in this section to evaluate the following integrals. $$\int \frac{x-2}{x^{2}+6 x+13} d x$$

Short Answer

Expert verified
Question: Evaluate the following integral: $$\int \frac{x-2}{x^2+6x+13} dx$$ Answer: $$\int \frac{x-2}{x^2+6x+13} dx = -\frac{5}{4} \ln|x+3| + 2 \arctan\left(\frac{x+3}{2}\right) + C$$

Step by step solution

01

Rewrite the integrand as a partial fraction decomposition

Write the integrand as a sum of two simpler fractions: $$\frac{x-2}{x^2+6x+13} = \frac{A}{x+3} + \frac{B}{(x+3)^2+4}$$
02

Find A and B

Clear the denominators by multiplying both sides of the equation by \((x^2+6x+13)\): $$(x-2) = A[(x+3)^2+4] + B(x+3)$$ Now, we can plug in some values of x to find A and B. By setting \(x=-3\), we get: $$-5 = A(4)$$ Solve for A: $$A = -\frac{5}{4}$$ Now, differentiate the right side with respect to x to find B: $$1 = 2A(x+3) + B$$ Insert the value of A we found earlier: $$1 = 2\left(-\frac{5}{4}\right)(x+3) + B$$ Now, again set \(x = -3\): $$1 = B$$ Now, we have \(A=-\frac{5}{4}\) and \(B=1\). So, we can rewrite the integrand as: $$\int \frac{-\frac{5}{4}}{x+3} + \frac{1}{(x+3)^2+4} dx$$
03

Perform a substitution

Substitute \({u} = x+3\) and \(du = dx\) to simplify the problem. The integral becomes: $$\int \frac{-\frac{5}{4}}{u} + \frac{1}{u^2+4} du$$
04

Split the integral

Separate the integral into two parts: $$-\frac{5}{4}\int \frac{1}{u} du + \int \frac{1}{u^2+4} du$$
05

Integrate each part

Integrate the first part: $$-\frac{5}{4} \ln|u| + C_1 = -\frac{5}{4} \ln|x+3| + C_1$$ For the second part, rewrite \(\int \frac{1}{u^2+4} du\) as \(\int \frac{1}{(\frac{u^2}{4})+(1)} du\). Now, substitute \({v} = \frac{u}{2}\), so \(dv = \frac{1}{2} du\). The integral becomes: $$\int \frac{1}{v^2+1} (2 dv) = 2 \int \frac{1}{v^2+1} dv = 2 \arctan(v) + C_2 = 2 \arctan\left(\frac{x+3}{2}\right) + C_2$$
06

Combine the integrals and simplify the result

Add the two integrals and their constants: $$-\frac{5}{4} \ln|x+3| + 2 \arctan\left(\frac{x+3}{2}\right) + C$$ So, the final solution is: $$\int \frac{x-2}{x^2+6x+13} dx = -\frac{5}{4} \ln|x+3| + 2 \arctan\left(\frac{x+3}{2}\right) + C$$

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

a. Verify the identity \(\sec x=\frac{\cos x}{1-\sin ^{2} x}\) b. Use the identity in part (a) to verify that \(\int \sec x d x=\frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\) (Source: The College Mathematics Joumal \(32,\) No. 5 (November 2001))

Graph the function \(f(x)=\frac{\sqrt{x^{2}-9}}{x}\) and consider the region bounded by the curve and the \(x\) -axis on \([-6,-3] .\) Then evaluate \(\int_{-6}^{-3} \frac{\sqrt{x^{2}-9}}{x} d x .\) Be sure the result is consistent with the graph.

By reduction formula 4 in Section 3 $$\int \sec ^{3} u d u=\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|)+C$$ Graph the following functions and find the area under the curve on the given interval. $$f(x)=\left(9-x^{2}\right)^{-2},\left[0, \frac{3}{2}\right]$$

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{e^{x}}{\left(e^{x}-1\right)\left(e^{x}+2\right)} d x$$

Use integration by parts to evaluate the following integrals. $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

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.