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Chapter 5: Problem 69

Calculate the integrals. $$ \int \frac{\exp (x)}{1+\exp (2 x)} d x $$

Short Answer

Expert verified
The integral is \( \arctan(\exp(x)) + C \).

Step by step solution

01

Let the Substitution Take Over

First, let's use a substitution to simplify the integral. Set \( u = \exp(x) \). Then the derivative of \( u \) with respect to \( x \) is \( du = \exp(x) \, dx \). This transforms our integral into \( \int \frac{1}{1 + u^2} \, du \).
02

Recognize the New Form

The integral \( \int \frac{1}{1 + u^2} \, du \) is a standard integral that corresponds to the arctangent function. This is a well-known result that computes to \( \arctan(u) + C \), where \( C \) is the constant of integration.
03

Reverse the Substitution

Replace \( u \) back with \( \exp(x) \) to complete the solution in terms of the original variable. Thus, we substitute \( u = \exp(x) \) back into the solution, resulting in \( \arctan(\exp(x)) + C \).

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Key Concepts

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

Substitution Method
In calculus, the substitution method is a powerful technique used to simplify complex integrals by introducing a new variable. This method essentially involves substituting a part of the integral with a single variable to reduce its complexity. In our example, we substitute \( u = \exp(x) \) for the given integral \( \int \frac{\exp(x)}{1+\exp(2x)} \, dx \).
  • First, notice the appearance of \( \exp(x) \) in both the numerator and denominator of the integrand. Setting \( u = \exp(x) \) simplifies the expression significantly.
  • To change the variable of integration from \( x \) to \( u \), find the derivative: \( du = \exp(x) \, dx \), which directly replaces \( \exp(x) \, dx \) in our integral.
  • Upon substitution, the integral becomes \( \int \frac{1}{1 + u^2} \, du \), which is much more manageable.
The beauty of the substitution method lies in its ability to transform an intricate problem into a recognizably simpler form, making it easier to integrate.
Arctangent Function
Upon transforming the integral using substitution, we encounter the integral \( \int \frac{1}{1+u^2} \, du \). This is a classical form that directly correlates with the arctangent function.
  • The integral \( \int \frac{1}{1+u^2} \, du \) equates to \( \arctan(u) + C \), where \( C \) is the integration constant.
  • The arctangent function, \( \arctan(x) \), is the inverse of the tangent function, and it maps real numbers to angles, specifically in radians.
  • In practical terms, when you see \( \frac{1}{1+u^2} \), it should trigger a recognition of the arctangent function as its antiderivative.
Understanding the connections between integral expressions and trigonometric functions can significantly streamline solving integrals and make these problems more intuitive.
Exponential Functions
Exponential functions, denoted as \( \exp(x) \) or \( e^x \), are fundamental in calculus and appear frequently in integration. They have unique properties that make them particularly interesting in mathematics.
  • The exponential function \( \exp(x) \) grows rapidly and is defined as its own derivative, meaning \( \frac{d}{dx}[\exp(x)] = \exp(x) \).
  • In the context of our integral \( \int \frac{\exp(x)}{1+\exp(2x)} \, dx \), \( \exp(x) \) appears in both the numerator and denominator, making it an ideal candidate for substitution.
  • The substitution simply replaces \( \exp(x) \) with \( u \), taking advantage of its derivative being \( \exp(x) \, dx \), simplifying integration considerably.
Grasping the characteristics of exponential functions can empower you to handle complex integrals more efficiently, and they are instrumental in various fields such as compound interest, population dynamics, and natural phenomena modeling.

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Most popular questions from this chapter

The graphs of \(y=x+2 / x\) and \(y=3+(x-2)^{2}\) intersect at \(P=(2,3),\) at one point to the left of \(P,\) and at one point to the right of \(P\). Find the total area of the regions that are between the two graphs.

An integral \(\int_{a}^{b} f(x) d x\) and a positive integer \(N\) are given. Compute the exact value of the integral, the Simpson's Rule approximation of order \(N,\) and the absolute error \(\varepsilon\). Then find a value \(c\) in the interval \((a, b)\) such that \(\varepsilon=(b-a)^{5}\left|f^{(4)}(c)\right| /\left(180 \cdot N^{4}\right) .\) (This form of the error, which resembles the Mean Value Theorem, implies inequality \((5.8 .4) .)\) $$ \int_{1}^{e} 1 / x d x \quad N=4 $$

Calculate the area of the region between the pair of curves. $$ y=(x-3) / 2 \quad x=y^{2} $$

Find the area of the region that is bounded by the graphs of \(y=f(x)\) and \(y=g(x)\) for \(x\) between the abscissas of the two points of intersection. $$ f(x)=x^{2}+5 \quad g(x)=2 x^{2}+1 $$

A function \(f\) is defined on a specified interval \(I=[a, b] .\) Calculate the area of the region that lies between the vertical lines \(x=a\) and \(x=b\) and between the graph of \(f\) and the \(x\) -axis. $$ f(x)=12-9 x-3 x^{2} \quad I=[-2,2] $$
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