Chapter 5: Problem 104
Finding an Indefinite Integral In Exercises \(91-108,\) find the indefinite integral. $$ \int \frac{2 e^{x}-2 e^{-x}}{\left(e^{x}+e^{-x}\right)^{2}} d x $$
Short Answer
Expert verified
The indefinite integral of \(\int \frac{2 e^{x}-2 e^{-x}}{(e^{x}+e^{-x})^{2}} dx\) is \( -2/e^{x} - 2/e^{-x} + C \)
Step by step solution
01
Define a new variable u
First, define a new variable \( u \) such that \( u = e^{x} + e^{-x} \). It simplifies the integral to render it more tractable by substituting \( u \) into the equation.
02
Calculate du (derivative of u)
Calculate the derivative of \( u \) with respect to \( x \), which is \( du/dx = e^{x} - e^{-x} \). Then multiply by \( dx \) which gives \( du = (e^{x} - e^{-x}) dx \).
03
Modify the function and apply the power rule for integration
Rearrange using u substitution we get, \(\int \frac{2 du}{u^{2}} \). This simplifies our integral and we now apply power rule which states \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \) with \( n = -2 \), we get \( -2u^{-1} \) plus a constant.
04
Substitute back for u
Finally, substitute the original variable \( u \) back into the equation. This yields our final answer: \( -2/e^{x} - 2/e^{-x} + C \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding U-Substitution
When it comes to solving indefinite integrals, one of the most helpful techniques is the method of u-substitution. This method is especially useful when an integral contains a composite function. The essence of u-substitution is to simplify the integral by substituting part of the integrand with a new variable, often called "u."
This technique transforms the complex integral into a simpler form that is much easier to evaluate. Here's how it works:
This technique transforms the complex integral into a simpler form that is much easier to evaluate. Here's how it works:
- Select a part of the expression in the integral that can be substituted with "u." This part is usually the inside function of a composite function.
- Express the derivative of "u" in terms of "x." This helps in replacing "dx" with "du."
- Rewrite the original integral in terms of "u" and "du."
Applying the Power Rule
One of the cornerstones of calculus integration is the power rule. The power rule is a basic formula that helps compute integrals of functions that are purely expressed as powers of x. It states that the indefinite integral of \(x^n\) with respect to x is \(\int x^n dx = \frac{x^{n+1}}{n+1} + C,\) where \(neq-1\).
In the context of our exercise, using u-substitution gives us \(\int \frac{2}{u^2} du.\) This integral can be directly solved using the power rule by recognizing that \(\frac{1}{u^2}\) is equivalent to \(u^{-2}.\)
By applying the power rule, the integral simplifies to \(-2u^{-1} + C,\) where \(C\) represents the integration constant. The power rule is a straightforward method when handling powers of a variable and significantly reduces computation effort.
In the context of our exercise, using u-substitution gives us \(\int \frac{2}{u^2} du.\) This integral can be directly solved using the power rule by recognizing that \(\frac{1}{u^2}\) is equivalent to \(u^{-2}.\)
By applying the power rule, the integral simplifies to \(-2u^{-1} + C,\) where \(C\) represents the integration constant. The power rule is a straightforward method when handling powers of a variable and significantly reduces computation effort.
The Essence of Calculus Integration
Calculus integration is a fundamental concept in calculus that deals with the notion of finding antiderivatives. An antiderivative is a function whose derivative is the given function. Indefinite integrals, unlike definite integrals, do not have specific limits and involve a constant of integration "C."
Integration reverses the process of differentiation and finds the original function given its derivative. It is a powerful tool in finding areas under curves, solutions to differential equations, and more.
Keys to successful integration:
Integration reverses the process of differentiation and finds the original function given its derivative. It is a powerful tool in finding areas under curves, solutions to differential equations, and more.
Keys to successful integration:
- Recognize patterns and standard forms of integrals.
- Use substitution methods when expressions are complex.
- Apply rules like the power rule, method of partial fractions, or trigonometric identities where applicable.
Variable Substitution Simplified
Variable substitution, often termed as u-substitution in calculus, is a strategic choice made to simplify the task of integration. By changing variables, we aim to change the way the integral looks, making it simpler and often reducing it to a well-known form that we can easily manage.
In the exercise, the use of variable substitution involved letting \(u = e^{x} + e^{-x}\). This effectively transformed the original complex integral into a form that was easier to handle: \(\int \frac{2}{u^2} du\).
Key advantages of variable substitution:
In the exercise, the use of variable substitution involved letting \(u = e^{x} + e^{-x}\). This effectively transformed the original complex integral into a form that was easier to handle: \(\int \frac{2}{u^2} du\).
Key advantages of variable substitution:
- Transforms difficult integrals into simpler, more manageable forms.
- Makes use of known integration rules and formulas to solve.
- Simplifies integration by breaking down expressions into simpler parts.