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Chapter 4: Problem 43

Find the indefinite integral. $$ \int \frac{e^{-x}}{1+e^{-x}} d x $$

Short Answer

Expert verified
The short answer to the indefinite integral \( \int \frac{e^{-x}}{1+e^{-x}} \, dx \) is: $$ -\ln |1+e^{-x}| + C $$

Step by step solution

01

Substitution

Let's substitute \(u = 1 + e^{-x}\). To find the derivative of \(u\) with respect to \(x\), we have: $$ \frac{du}{dx} = -e^{-x}, $$ Now, we can solve for \(dx\): $$ dx = \frac{du}{-e^{-x}} $$ Replacing the expression in the integral, we have: $$ \int \frac{e^{-x}}{u} \, \frac{du}{-e^{-x}} $$ Notice that the term \(e^{-x}\) cancels out: $$ \int \frac{1}{u} \, (-du) $$
02

Find the antiderivative

Now, we can find the antiderivative of the simplified integrand: $$ -\int \frac{1}{u} \, du $$ The antiderivative of \( \frac{1}{u} \) is \( \ln |u| \), so we have: $$ -\ln |u| + C $$ where \(C\) is the constant of integration.
03

Substitute back for x

Now, we replace \(u\) with the original substitution: $$ -\ln |1+e^{-x}| + C $$ This is the indefinite integral: $$ \int \frac{e^{-x}}{1+e^{-x}} \, dx = -\ln |1+e^{-x}| + C $$

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Key Concepts

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

Substitution Method
In solving integrals, especially those that appear complex at first glance, the substitution method is a powerful tool. This technique involves replacing a part of the integrand with a new variable, usually denoted byThis assists in simplifying the integral, making it easier to solve.

In this exercise, we begin by setting the substitution
This substitution allows us to transform the original integral expression into a simpler form: \( \int \frac{1}{u} (-du) \). Notice how the complex expression becomes much more manageable, thanks to the cancellation of \( e^{-x} \). The substitution method not only aids in handling difficult integrals but also facilitates focusing on the key elements of the function.
Antiderivative
Having made the expression simpler by substitution, the next step is to find the antiderivative. The antiderivative, or the "indefinite integral," of a function f(x) is a function F(x), such that the derivative of F(x) is equal to f(x).

In this case, our new integral is \( \int \frac{1}{u} (-du) \). The antiderivative of \( \frac{1}{u} \) is a logarithmic function, specifically \( \ln |u| \). Therefore, integrating with respect to \( u \), we find:
  • \( -\ln |u| \)

This result highlights the beauty and simplicity of the natural logarithm in integration problems. Calculating the antiderivative converts the differential posit into its continuous form, which is crucial to solving the integral.
Constant of Integration
Once the antiderivative is determined, it is accompanied by a constant of integration, denoted usually by C. This constant represents any constant value that could have been differentiated to zero in the original function.

In our scenario, when integrating \( -\ln |u| \) with respect to \( u \), we add C to signify that there are infinitely many possible solutions vertically shifted on the graph.

Moving from the substitution variable back to the original, we replace \( u \) with \( 1 + e^{-x} \) to return to the x-variable domain:
  • \( -\ln |1+e^{-x}| + C \)

This inclusion is vital in indefinite integrals, setting them apart from definite integrals where specific bounds eliminate the arbitrary constant. Understanding the constant of integration ensures you see the full solution set of antiderivatives, maintaining the fundamental property of indefinite integrals.

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Most popular questions from this chapter

Express the area of the region under the graph of the function f over the interval as the limit of a sum (use the right endpoints), (b) use a computer algebra system (CAS) to find the sum obtained in part (a) in compact form, and (c) evaluate the limit of the sum found in part (b) to obtain the exact area of the region. $$ f(x)=\sin x ; \quad\left[0, \frac{\pi}{2}\right] $$

In exercise, (a) find the number \(c\) whose existence is guaranteed by the Mean Value Theorem for Integrals for the function \(f\) on \([a, b]\), and (b) sketch the graph of f on \([a, b]\) and the rectangle with base on \([a, b]\) that has the same area as that of the region under the graph of \(f\). $$ f(x)=x^{3} ; \quad[0,2] $$

Express the area of the region under the graph of the function f over the interval as the limit of a sum (use the right endpoints), (b) use a computer algebra system (CAS) to find the sum obtained in part (a) in compact form, and (c) evaluate the limit of the sum found in part (b) to obtain the exact area of the region. $$ f(x)=x^{4} ; \quad[0,2] $$

Prove that $$ \int_{-1 / 2}^{1 / 2} 2^{\cos x} d x=2 \int_{0}^{1 / 2} 2^{\cos x} d x $$

Velocity of an Attack Submarine The following data give the velocity of an attack submarine taken at 10 -min intervals during a submerged trial run. $$\begin{array}{|l|ccccccc|} \hline \text { Time } t \text { (hr) } & 0 & \frac{1}{6} & \frac{1}{3} & \frac{1}{2} & \frac{2}{3} & \frac{5}{6} & 1 \\ \hline \text { Velocity } v \text { (mph) } & 14.2 & 24.3 & 40.2 & 45.0 & 38.5 & 27.6 & 12.8 \\ \hline \end{array}$$ Use Simpson's Rule to estimate the distance traveled by the submarine during the 1 -hr submerged trial run.
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