Problem 46 - Essential Calculus

Question

Find the indefinite integral. $$ \int \frac{2 e^{x}-2 e^{-x}}{\left(e^{x}+e^{-x}\right)^{2}} d x $$

Step-by-Step Solution

Answer

\(-2/(e^x + e^{-x}) + C\

Step 1: Identify the nucleic formula

The nucleic formula for integrals is \(\int f'(x) / f(x) dx = ln |f(x)| + C\). When the derivative of a function \(f(x)\) is divided by \(f(x)\) itself, the integral can be computed as the natural log of the absolute value of \(f(x)\).

Step 2: Perform the substitution

Make the substitution \(u = e^x + e^{-x}\). Then compute the differential \(du\) by taking the derivative of both sides (with respect to \(x\)): \(du = e^x - e^{-x} dx\). It is noticed that \(2du = 2e^x - 2e^{-x} dx\). This simplifies the given integral to \(\int \frac{2du}{u^2}\).

Step 3: Use the power rule for integration

For the integral of the form \(\int u^n du\), where \(n \neq -1\), the power rule for integration states that \(\int u^n du = \frac{1}{n+1} u^{n+1} + C\). Applying the power rule to \(\int \frac{2}{u^2} du = \int 2u^{-2} du\), we get \(-2u^{-1} + C\).

Step 4: Substitute back and simplify

Substitute \(u = e^x + e^{-x}\) back into the integral to get the final result: \(-2/(e^x + e^{-x}) + C\).

Key Concepts

Integration by Substitution

Integration by substitution is a powerful tool for solving complex integrals - it's akin to untangling a knot by carefully pulling on the right strings. Picture it like finding a key that unlocks a treasure chest; once you identify this key, the inside becomes accessible and straightforward. When faced with an integral that does not fit the standard forms, you can perform a substitution to simplify it. Often, this involves introducing a new variable, say \(u\), that transforms the integral into a simpler form.

Using the substitution\(u = e^x + e^{-x}\) as in the given problem, you're essentially looking for a pattern in the integrand that hints at a simpler underlying structure. Once you've made the substitution and calculated the new differential \(du\), the integral becomes much easier to manage. It's similar to changing the language of a difficult text into one you understand better - the content remains the same, but it's conveyed in terms you can easily work with.

Power Rule for Integration

The power rule for integration is like a trusty hammer in your mathematics toolkit - it's essential and reliably effective for certain types of jobs. When you're dealing with a function in the form \( u^n \), where \(n\) is a real number that isn't -1, you can use the power rule which says \( \int u^n du = \frac{1}{n+1} u^{n+1} + C \). In our problem, after the substitution process, you have \( 2u^{-2} \), and applying the power rule here is straightforward. You end up with \(-2u^{-1} + C\), which is the antiderivative of the function. It's like applying a simple recipe to bake a specific type of bread – knowing what ingredients and amounts to mix will get you the desired result every time.

Understanding when and how to apply this rule is key in becoming proficient in integrating a wide variety of functions. It's a logical step that follows the process of substitution and further simplifies the expression into something that's easily solvable.

Natural Logarithm

When the groove of a function aligns perfectly with the pattern of its derivative, the natural logarithm often plays a star role in the solution. It is elegantly captured in the formula \( \int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + C \), which is exceptionally useful when you're confronted with a quotient of a function and its derivative, as we saw in the exercise. Just as the integral of the constant function 1 is the identity function \( x \), the natural logarithm \( \ln \) serves as the antiderivative for the reciprocal function \( \frac{1}{x} \), reflecting a deep symmetry in the landscape of mathematical functions.

Integration sometimes leads to a natural log when a function's inner structure tempts you down that path, as with the nucleic formula encountered in the initial solution. In our exercise, after substitution and the application of the power rule, you discover that the expression \(-2u^{-1} + C\) also confirms this close relationship. The result encapsulates the beauty of integration when it leads to the natural logarithm, offering a moment of clarity where the intricate tapestry of calculus weaves together.