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Prove that the function \(f(x)=e^{x} /\left(e^{5 x}+e^{x}+1\right)\) has an elementary primitive. (Do not try to find it!)

Short Answer

Expert verified
The function \( f(x) = \frac{e^x}{e^{5x} + e^x + 1} \) has an elementary primitive by rational function integrability post substitution.

Step by step solution

01

- Identify the type of function

The given function is rational of the form \( f(x) = \frac{e^x}{e^{5x} + e^x + 1} \). Here, both the numerator and the denominator are functions of the exponential form.
02

- Simplify the function

Simplify \( f(x) \) by attempting to factor common terms in the denominator. Notice that \( e^{5x} \) is much larger than \( e^x \), so consider common exponential factors.
03

- Use substitution for simplification

Let's use the substitution \( u = e^x \). Then, \( du = e^x dx = u dx \). The function transforms as \( f(u) = \frac{u}{u^5 + u + 1} \). This simplifies and uses typical rational function integrals.
04

- Determine the type of integral

Recognize that this function fits the forms that are typically solvable by elementary functions. Specifically, the rational function after substitution suggests that standard calculus techniques can be used.
05

- Conclude with the integrability

Given that the resulting substitution leads to a rational function appeal to the fact that rational functions with polynomial terms in the forms we encountered generally possess elementary primitives.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Function
A rational function is a type of function that is represented as the ratio of two polynomials. In other words, it is in the form \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. For the given problem, the function \( f(x) = \frac{e^x}{e^{5x} + e^x + 1} \) is a rational function. Though it might seem complicated due to the presence of exponentials, it fundamentally fits the form of a rational function once we make the proper substitution. Understanding rational functions helps simplify and manage complex integrals more by breaking them down into more familiar polynomial operations.
Recognizing this structure is the first step in solving the integral, as shown in step 1 of the solution. Step 1 clarifies that although the numerator and denominator are not simple polynomials, they follow a similar behavior due to their exponential nature.
Exponential Function
An exponential function is one where the variable appears in the exponent. The general form for an exponential function is \( f(x) = a^x \), where \( a \) is a constant. In the original problem, both the numerator \( e^x \) and the terms of the denominator \( e^{5x} + e^x + 1 \) are exponential functions. Exponentials often grow at an extremely rapid rate, particularly as the exponent increases. Understanding how these functions behave is crucial in simplifying and solving integrals. In step 2, the solution shows the importance of analyzing the growth rates—since \( e^{5x} \) grows much faster than \( e^x \, \) it can dominate the behavior of the denominator. Recognizing this behavior helps in simplifying the rational function and making the substitution process clearer.
Such understanding provides insight into why and how we proceed with substitution in the next steps.
Substitution Method
The substitution method is a fundamental technique in calculus used for simplifying integrals. It involves changing the variable of integration to transform the integral into a simpler form. For the given function, we use substitution to transform the exponential forms into a rational function. In step 3 of the solution, the substitution \( u = e^x \) simplifies the given problem significantly. With \( du = e^x \ dx = u \ dx \, \) we can rewrite the integral with respect to \( u \). The function then becomes \( f(u) = \frac{u}{u^5 + u + 1} \, \) making it easier to handle. This rational form simplifies the process of finding the integral, as standard techniques for rational functions can now be applied.
Substitution is particularly useful when dealing with exponential functions since it can turn complex expressions into simpler polynomial-like forms.
Integral Simplification
Integral simplification involves using various techniques to rewrite a complex integral into a more manageable form. This might involve algebraic manipulation, substitution, or recognizing patterns among integrals. In the final steps of the solution (steps 4 and 5), the problem confirms that the function indeed fits into a framework that allows it to be integrated using standard methods. After substitution, the function \( \frac{u}{u^5 + u + 1} \) is recognized as a rational function that can typically be integrated using known techniques.
Given that rational functions made up of polynomials usually possess elementary primitives, the fundamental theorem of calculus ensures integrability. Thus, understanding this principle concludes our problem: the original function does indeed have an elementary primitive, as expected.
Simplifying integrals effectively demonstrates not just an understanding of individual techniques but also an appreciation for how different mathematical concepts interconnect to solve problems.

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Most popular questions from this chapter

This problem contains some integrals which require little more than algebraic manipulation, and consequently test your ability to discover algebraic tricks, rather than your understanding of the integration processes. Nevertheless, any one of these tricks might be an important preliminary step in an honest integration problem. Moreover, you want to have some feel for which integrals are easy, so that you can see when the end of an integration process is in sight. The answer section, if you resort to it, will only reveal what algebra you should have used. (i) \(\int \frac{\sqrt[5]{x^{3}}+\sqrt[6]{x}}{\sqrt{x}} d x\). (ii) \(\int \frac{d x}{\sqrt{x-1}+\sqrt{x+1}}\). (iii) \(\int \frac{e^{x}+e^{2 x}+e^{3 x}}{e^{4 x}} d x\). (iv) \(\int \frac{a^{x}}{b^{x}} d x\). (v) \(\int \tan ^{2} x d x.\) (Trigonometric integrals are always very touchy, because there are so many trigonometric identities that an easy problem can easily look hard.) (vi) \(\int \frac{d x}{a^{2}+x^{2}}\). (vii) \(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}\). (viii) \(\int \frac{d x}{1+\sin x}\). (ix) \(\int \frac{8 x^{2}+6 x+4}{x+1} d x\). (x) \(\int \frac{1}{\sqrt{2 x-x^{2}}} d x\).

The following integrations involve substitutions of various types. There is no substitute for cleverness, but there is a general rule to follow: substitute for an expression which appears frequently or prominently; if two different troublesome expressions appear, try to express them both in terms of some new expression. And don't forget that it usually helps to express \(x\) directly in terms of \(u,\) to find out the proper expression to substitute for \(d x.\) (i) \(\int \frac{d x}{1+\sqrt{x+1}}.\) (ii) \(\int \frac{d x}{1+e^{x}}.\) (iii) \(\int \frac{d x}{\sqrt{x}+\sqrt[3]{x}}.\) (iv) \(\int \frac{d x}{\sqrt{1+e^{x}}}.\) (The substitution \(u=e^{x}\) leads to an integral requiring yet another substitution; this is all right, but both substitutions can be done at once.) (v) \(\int \frac{d x}{2+\tan x}.\) (vi) \(\int \frac{d x}{\sqrt{\sqrt{x}+1}}.\) (Another place where one substitution can be made to do the work of two.) (vii) \(\int \frac{4^{x}+1}{2^{x}+1} d x.\) (viii) \(\int e^{\sqrt{x}} d x\) (ix) \(\int \frac{\sqrt{1-x}}{1-\sqrt{x}} d x.\) (In this case two successive substitutions work out best; there are two obvious candidates for the first substitution, and either will work.) *(x) \(\int \sqrt{\frac{x-1}{x+1}} \cdot \frac{1}{x^{2}} d x.\)

Express \(\int \log (\log x) d x\) in terms of \(\, \int(\log x)^{-1} d x .\) (Neither is expressible in terms of elementary functions.)

Let \(c\) be a curve represented parametrically by \(u\) and \(v\) on \([a, b],\) and let \(h\) be an increasing function with \(h(\bar{a})=a\) and \(h(\bar{b})=b .\) Then on \([\bar{a} . \bar{b}]\) the functions \(\bar{u}=u\) o \(h, \bar{v}=v\) o \(h\) give a parametric representation of another curve \(\bar{c} ;\) intuitively, \(\bar{c}\) is just the same curve \(c\) traversed at a different rate. (a) Show, directly from the definition of length, that the length of \(c\) on \([a, b]\) equals the length of \(\bar{c}\) on \([\bar{a}, \bar{b}].\) (b) Assuming differentiability of any functions required, show that the lengths are equal by using the integral formula for length, and the appropriate substitution.

Derive the formula for \(\int \sec x \, d x\) in the following two ways: (a) By writing $$\begin{aligned} \frac{1}{\cos x} &=\frac{\cos x}{\cos ^{2} x} \\ &=\frac{\cos x}{1-\sin ^{2} x} \\ &=\frac{1}{2}\left[\frac{\cos x}{1+\sin x}+\frac{\cos x}{1-\sin x}\right], \end{aligned}$$ an expression obviously inspired by partial fraction decompositions. Be sure to note that \(\int \cos x /(1-\sin x) d x=-\log (1-\sin x) ;\) the minus sign is very important. And remember that \(\frac{1}{2} \log \alpha=\log \sqrt{\alpha} .\) From there on, keep doing algebra, and trust to luck. (b) By using the substitution \(t=\tan x / 2 .\) One again, quite a bit of manipulation is required to put the answer in the desired form; the expression \(\tan x / 2\) can be attacked by using Problem \(15-9,\) or both answers can be expressed in terms of \(t .\) There is another expression for \(\int \sec x \, d x\) which is less cumbersome than \(\log (\sec x+\tan x) ;\) using Problem 15-9 we obtain $$\int \sec x \, d x=\log \left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)=\log \left(\tan \left(\frac{x}{2}+\frac{\pi}{4}\right)\right).$$ This last expression was actually the one first discovered. and was due, not to any mathematician's cleverness, but to a curious historical accident: In 1599 Wright computed nautical tables that amounted to definite integrals of sec. When the first tables for the logarithms of tangents were produced, the correspondence between the two tables was immediately noticed (but remained unexplained until the invention of calculus).

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