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Chapter 5: Problem 35

Evaluate the integrals using appropriate substitutions. $$\int \frac{e^{x}}{1+e^{2 x}} d x$$

Short Answer

Expert verified
\( \tan^{-1}(e^{x}) + C \)

Step by step solution

01

Set up the substitution

To simplify the integral, let's use a substitution. Let \( u = e^{x} \). This means the derivative with respect to \( x \) is \( \frac{du}{dx} = e^{x} \), or \( du = e^{x} \, dx \).
02

Substitute into the integral

Substitute \( u = e^{x} \) and \( dx = \frac{du}{e^{x}} \) into the integral. The original integral \( \int \frac{e^{x}}{1 + e^{2x}} \, dx \) becomes \( \int \frac{e^{x}}{1 + u^2} \cdot \frac{du}{e^{x}} \). The \( e^{x} \) terms cancel, simplifying to \( \int \frac{1}{1+u^2} \, du \).
03

Integrate using a known integral formula

The integral \( \int \frac{1}{1+u^2} \, du \) is a standard form, known to be \( \tan^{-1}(u) + C \), where \( C \) is the constant of integration.
04

Substitute back to original variable

Replace \( u \) with \( e^{x} \) to obtain the result in terms of \( x \). This gives \( \tan^{-1}(e^{x}) + C \).

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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 technique in integration, often used when an integral contains a function and its derivative. The goal is to simplify the integral to a more recognizable form. In the given exercise, we started by setting a substitution: let \( u = e^x \). This means the derivative \( du = e^x \, dx \), which is perfectly aligned to replace the terms in the original integral.
  • By introducing \( u \), we turn the complex integral \( \int \frac{e^x}{1+e^{2x}} \, dx \) into a simpler one, specifically \( \int \frac{1}{1+u^2} \, du \).
  • This process of substitution simplifies the calculation by changing the variable from \( x \) to \( u \).
Utilizing substitutions is essential, especially when dealing with integrals involving compositions of functions. With this method, we can often transform an intimidating problem into a familiar and straightforward form.
Standard Integral Formulas
Standard integral formulas serve as shortcuts, allowing for quick evaluation of integrals once a recognition of the pattern is made. In this problem, after using substitution to simplify the integral, we recognized a standard form:
  • The integral \( \int \frac{1}{1+u^2} \, du \) is a well-known formula leading to the result \( \tan^{-1}(u) + C \).
  • These standard forms are typically memorized or listed in mathematical tables as they appear frequently across various problems.
Identifying and utilizing these formulas reduces complexity and computation time. They provide a bridge between complex functional forms and simple results, like recognizing \( \int \frac{1}{1+x^2} \) leads directly to an inverse tangent function. By matching the simplified integral to a standard form, we can swiftly conclude the integration process, leaving only the back-substitution to perform.
Inverse Trigonometric Functions
Inverse trigonometric functions, such as \( \tan^{-1}(u) \), play an essential role in integration, especially with rational functions like \( \frac{1}{1+u^2} \). This function specifically arises from considering the geometry of a right triangle where tangent and its inverse relationships are found.
  • The integral \( \int \frac{1}{1+u^2} \, du \) corresponds to the inverse tangent function \( \tan^{-1}(u) \), derived from the understanding of derivatives of the inverse trigonometric functions.
  • These functions are particularly useful in integrals involving quadratic expressions, helping to simplify results in a form that's both recognizable and usable in further calculations.
Once the integral is evaluated to \( \tan^{-1}(u) \), the final result remains incomplete without substituting back to the original variable \( x \). Thus, it's critical to retrace to \( \tan^{-1}(e^x) + C \), ensuring the solution consistently relates back to the problem's original terms and expressions.

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Most popular questions from this chapter

(a) Over what open interval does the formula $$ F(x)=\int_{0}^{x} \sec t d t $$ represent an antiderivative of \(f(x)=\sec x ?\) (b) Find a point where the graph of \(F\) crosses the \(x\) -axis.

$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$\sum_{k=1}^{100}\left(2^{k+1}-2^{k}\right)$$

It can be shown that every interval contains both rational and irrational numbers. Accepting this to be so, do you believe that the function $$ f(x)=\left\\{\begin{array}{lll} 1 & \text { if } & x \text { is rational } \\ 0 & \text { if } & x \text { is irrational } \end{array}\right. $$ is integrable on a closed interval \([a, b] ?\) Explain your reasoning.

Sketch the curve and find the total area between the curve and the given interval on the \(x\) -axis. $$y=e^{x}-1 ;[-1,1]$$

(a) Show that the area under the graph of \(y=x^{3}\) and over the interval \([0, b]\) is \(b^{4} / 4\) (b) Find a formula for the area under \(y=x^{3}\) over the interval \([a, b],\) where \(a \geq 0\)
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