Problem 68
Question
Use a change of variables to evaluate the following integrals. $$\int \frac{e^{2 x}}{e^{2 x}+1} d x$$
Step-by-Step Solution
Verified Answer
Answer: Use the substitution method with $u = e^{2x} + 1$. The solution to the integral is $\frac{1}{2} \ln{|e^{2x} + 1|} + C$.
1Step 1: Choose an appropriate substitution
Let's let a new variable u be:
$$u = e^{2x} + 1$$
2Step 2: Differentiate the substitution with respect to x
Differentiate u with respect to x:
$$\frac{du}{dx} = 2 e^{2x}$$
Rearrange this equation to get dx:
$$dx = \frac{1}{2e^{2x}}du$$
3Step 3: Substitute the new variable into the integral
Now, we will replace the variables in the integral with our new variable u and du:
$$\int \frac{e^{2 x}}{u} \frac{1}{2e^{2x}} du$$
Simplify the integral expression:
$$\frac{1}{2} \int \frac{1}{u} du$$
4Step 4: Evaluate the integral
This integral is now a basic integration problem:
$$\frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln{|u|} + C$$
5Step 5: Substitute the original variables back in
Replace u with the original expression in terms of x:
$$\frac{1}{2} \ln{|e^{2x} + 1|} + C$$
This is the final answer for the integral.
Key Concepts
change of variablesdefinite integralsindefinite integrals
change of variables
The change of variables is a pivotal concept in calculus, used to simplify complex integrals through substitution. When directly integrating a function seems challenging, we can introduce a new variable to rewrite the integral in a simpler form.
- In the case of our original function, it's given as \( \int \frac{e^{2x}}{e^{2x}+1} \, dx \). Here, a direct approach to solving seems difficult due to the complexity in the exponent.
- To apply change of variables or substitution, select a part of the integral expression to act as your new variable \(u\). This choice often stems from identifying a nested function or composite part of the given function. Here, choosing \( u = e^{2x} + 1 \) streamlines the integration process.
- It requires differentiating this \( u\) with respect to \( x\) and finding \(dx\) in terms of \(du\). In this case, we get \( \frac{du}{dx} = 2e^{2x} \) which simplifies to \( dx = \frac{1}{2e^{2x}} du \).
definite integrals
Definite integrals refer to the evaluation of the integral of a function over a specified interval. Unlike indefinite integrals, they yield a real number that represents the net area under the curve of a function.
- In practice, after performing a suitable substitution or integration directly, definite integrals require you to use the limits of integration, applying them to the antiderivative.
- The outcome is a numerical value that may represent a significant physical quantity such as area, volume, or even total change in certain applications.
- When using a change of variables approach, it's vital to adjust the limits from the original ones based on \( x \) to the transformed limits in terms of \( u \), using the substitution. This ensures the integration process remains valid and accurate.
indefinite integrals
Indefinite integrals find the antiderivative of a function, returning a general formula that includes a constant of integration, \(C\). The outcome is not a specific value but a family of functions.
- For the integral \( \frac{1}{2} \int \frac{1}{u} \, du \), it yields an antiderivative of the form \( \frac{1}{2} \ln{|u|} + C \).
- This solution inherently includes \( + C\) to account for the indefinite nature of antiderivatives since differentiating it must yield the original function through all potential vertical shifts of the function graph.
- In practical terms, the expression \( \frac{1}{2} \ln{|e^{2x} + 1|} + C \) is the general solution before applying any boundary conditions or added specificity that might arise in definite contexts.
Other exercises in this chapter
Problem 67
Simplify the following expressions. $$\frac{d}{d x} \int_{-x}^{x} \sqrt{1+t^{2}} d t$$
View solution Problem 67
Consider two functions \(f\) and \(g\) on [1,6] such that \(\int_{1}^{6} f(x) d x=10, \int_{1}^{6} g(x) d x=5\) \(\int_{4}^{6} f(x) d x=5,\) and \(\int_{1}^{4}
View solution Problem 68
Suppose \(f\) is continuous on \([a, b]\) with \(f^{\prime \prime}(x)>0\) on the interval. It can be shown that $$(b-a) f\left(\frac{a+b}{2}\right) \leq \int_{a
View solution Problem 68
Simplify the following expressions. $$\frac{d}{d x} \int_{e^{x}}^{e^{2 x}} \ln t^{2} d t$$
View solution