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Chapter 6: Problem 17

Evaluate the following integrals. Include absolute values only when needed. \(\int \frac{e^{2 x}}{4+e^{2 x}} d x\)

Short Answer

Expert verified
Question: Evaluate the integral \(\int \frac{e^{2x}}{4+e^{2x}} dx\). Answer: \(-\arctan(\frac{e^{2x}}{2}) + C\)

Step by step solution

01

Substitute variable

Let's substitute \(u = e^{2x}\). To calculate \(du\), take the derivative with respect to x on both sides: \(du = \frac{d}{dx}(e^{2x}) dx\). Using the chain rule, we get \(du = 2 e^{2x} dx\). Now, let's rewrite the integral \(\int \frac{e^{2 x}}{4+e^{2 x}} d x\) in terms of \(u\): \(\int \frac{e^{2 x}}{4+e^{2 x}} d x = \int \frac{1}{4+u} (\frac{1}{2} du)\).
02

Evaluate the integral in terms of \(u\)

Now we need to evaluate the integral: \(\frac{1}{2} \int \frac{1}{4+u} du\). This integral can be solved by recognizing the integration as the form of arctangent. The Integral \(-\frac{1}{2} \int \frac{1}{4-u^2}(-du)\) is exactly arctangent integral. Now make a new substitution \(t = \frac{u}{2}\). Then we get \(dt = \frac{1}{2}du\). Thus our integral becomes: \(-\int \frac{1}{4-t^2} dt\). The result is an arctangent: \(-\arctan(t) + C\).
03

Substitute back the variables and simplify

Now let's substitute \(t\) back in terms of \(u\): \(-\arctan(\frac{u}{2}) + C = -\arctan(\frac{e^{2x}}{2}) + C\). So, the final result is: \(\int \frac{e^{2 x}}{4+e^{2 x}} d x = -\arctan(\frac{e^{2x}}{2}) + C\).

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Key Concepts

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

Substitution Method
The substitution method is a powerful tool for integrating complex functions. It simplifies an integral by changing variables, making the integral easier to evaluate. In this exercise, we begin by letting \(u = e^{2x}\). This transformation helps rewrite the integral in terms of a new variable \(u\).

This substitution effectively changes
  • original variable \(x\) into a simpler form \(u\).
  • complicated parts of the integral into straightforward expressions with \(u\).
To complete the substitution, we find \(du\) by differentiating \(e^{2x}\) (using rules of differentiation) and get \(du = 2e^{2x}dx\). This step is crucial as it aligns the integrand with the new variable \(u\). Through substitution, complex problems often reveal much simpler patterns.
Chain Rule
The chain rule is essential in calculus for differentiating compositions of functions. By recognizing how functions nest within each other, we can efficiently find derivatives. In this exercise, it was necessary for calculating \(du\).

Here's how the chain rule applies:
  • Given \(e^{2x}\), we treat \(2x\) as the inner function and \(e^x\) as the outer function.
  • Differentiating the outer function \(e^x\) gives \(e^{2x}\), and the derivative of the inner function \(2x\) is \(2\).
Multiplying these derivatives together gives \(2e^{2x}\), which is used for \(du\). Understanding and applying the chain rule is vital for these variable transformations.
Arctangent Integration
Arctangent integration comes into play when dealing with integrals of the form \(\frac{1}{a^2 + u^2}\). This type of integration leads to the use of arctangent functions. In our problem, after substitution, we end up with an integral that matches this structure.

Steps involved include:
  • Recognizing the integral format fitting arctangent rules.
  • Using the trigonometric identity for arctangent integration: \(\int \frac{1}{a^2 + u^2} \, du = \frac{1}{a}\arctan\left(\frac{u}{a}\right) + C\).
In our example, an intermediate substitution was key in turning the integral into a recognizable arctangent formula. Understanding these identities allows for streamlined integration when facing similar problems.
Definite and Indefinite Integrals
When performing integration, we distinguish between definite and indefinite integrals. Indefinite integrals provide a general form of antiderivatives, including a constant of integration \(C\). This exercise focuses on indefinite integrals.

Key differences include:
  • Indefinite integrals: Represent families of functions and are denoted without bounds. An example is \(\int f(x) \, dx = F(x) + C\).
  • Definite integrals: Calculate the net area under curves within specific limits. Unlike their indefinite counterparts, they yield a numerical result: \(\int_a^b f(x) \, dx\).
In our exercise, the solution results in an indefinite integral, showcasing understanding of fundamental integration principles. Recognizing the type of integral helps determine both approach and context for integration tasks.

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Most popular questions from this chapter

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