Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Question

Answered step-by-step

Evaluate the integral.

$ \displaystyle \int \frac{5x + 1}{(2x + 1)(x - 1)}\ dx $

Video Answer

Solved by verified expert

This problem has been solved!

Try Numerade free for 7 days

Like

Report

Official textbook answer

Video by J Hardin

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Missouri State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

02:15

Evaluate the integral.…

01:03

Evaluate the definite inte…

01:50

01:33

03:06

Evaluate the integrals.

02:47

02:05

Let's evaluate the following integral and let's go ahead and take the fraction inside the inte grant and go ahead and rewrite this using partial fraction decomposition. This is what the textbook would call case one. We have distinct linear factors in the denominator, so we can rewrite. This is a over two X plus one plus B over X minus one and then let's go ahead and multiply both sides by the denominator two X plus one x minus one. So we multiply on the left, the two X plus water and the X minus ones will cancel. So we're left with five x plus one on the right, where we multiply by this fraction by this product of linear polynomial. The two x plus ones will cancel when we multiply A. So we have a X minus one and then similarly for B, the X minus ones will cancel. And so we left over with two X plus one. So let's go ahead and rewrite the right hand side. So let's factor out an X here, and then we have a and then to be, and then we have B minus a. So if these two expressions are equal, then that must mean that the terms in front of the X term are equal and also that the constant terms are equal. So this gives us two equations. We have a plus to be equals five. But then we also have that B minus A is one. So we have a two by two system for A and B. Let's go ahead and solve that many ways To solve this one way is to just take this equation over here in green, solid for B and then go ahead and plug this be value into the other equation. So if we do that a plus to be replace, be with one plus a this is equal to five. But then we can go ahead and simplify this. We have three a equals five minus two. So that means a is one. And then using this equation over here, we also get that B as to so I'm running out of room here, let me go into the next page. So now the next thing, actually, before that what we should do here is go ahead and replace a and B with the values that we found. So now We'll replace A and B with one and two and then we'll integrate. Instead of integrating the left hand side, we'll go ahead and integrate the right hand side. So we have a was equal to one and then be was equal to two and now we have two. Integral to deal with. It may help you here to use a U sub. If you'd like to use the use up here, you can let you be two x plus one for this integral here. If you don't like that minus one on the X, you can go ahead and do another sub here. Let's do w this time then D w equals d X do u equals two d X so you can go ahead and carry out the U substitution here to evaluate this Integral. In either case, we are going to see natural algorithm. So for the first in overall, we get the one half natural log Absolute value two x plus one and for the second, integral to natural log Absolute value, X minus one plus our constancy. And that's our answer

View More Answers From This Book

Find Another Textbook

04:34

Consider the situation of a mass layoff (i.e., a factory shuts down) where 1…

03:55

Since the 3 -space is spanned by the three unit vectors defined in $(4.7),$ …

03:11

In the three-dimensional Euclidean space, what is the distance between the f…

03:01

Given the function $q=\left(v^{2}+v-56\right) /(v-7),(v \neq 7),$ find the l…

01:37

Given the total-cost function $C=Q^{3}-5 Q^{2}+12 Q+75$, write out a variabl…

01:35

Given the following four matrices, test whether any one of them is the inver…

04:11

On the basis of $(7.18),$ find the partial derivatives $\partial Y^{*} / \pa…

04:14

Are the following functions quasiconcave? Strictly so? First check graphical…

01:47

Let the demand and supply be $Q_{d}=\alpha-\beta P+\sigma \frac{d P}{d t} \q…

Find the product matrices in the following (in each case, append beneath eve…