Problem 53 - Calculus Single Variable

Question

In each of Exercises \(53-56,\) make a substitution before applying the method of partial fractions to calculate the given integral. $$ \int \frac{\exp (x)}{\exp (2 x)-1} d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{2} \ln \left| \frac{\exp(x) - 1}{\exp(x) + 1} \right| + C \).

Step 1: Identify a substitution

To simplify the integral, let's make the substitution that can ease working with the exponential functions. Use the substitution \( u = \exp(x) \). This gives us \( du = \exp(x) \, dx \), or equivalently \( dx = \frac{du}{u} \).

Step 2: Change the variable in the integral

With the substitution \( u = \exp(x) \), rewrite the integral in terms of \( u \). The integral becomes \( \int \frac{u}{u^2 - 1} \cdot \frac{du}{u} \), which simplifies to \( \int \frac{du}{u^2 - 1} \).

Step 3: Apply partial fraction decomposition

Recognize that \( u^2 - 1 \) can be factored as \( (u-1)(u+1) \). So, we use partial fraction decomposition: \( \frac{1}{(u-1)(u+1)} = \frac{A}{u-1} + \frac{B}{u+1} \).

Step 4: Solve for partial fraction coefficients

Set up the equation \( 1 = A(u+1) + B(u-1) \). Expand and collect terms to get \( 1 = (A+B)u + (A-B) \). For this to be true for all \( u \), \( A+B = 0 \) and \( A-B = 1 \). Solving these equations gives \( A = \frac{1}{2} \) and \( B = -\frac{1}{2} \).

Step 5: Write integral with partial fractions

Substitute the values of \( A \) and \( B \) into the partial fraction decomposition: \( \frac{1}{u^2 - 1} = \frac{1/2}{u-1} - \frac{1/2}{u+1} \), and rewrite the integral: \( \int \frac{1/2}{u-1} \, du - \int \frac{1/2}{u+1} \, du \).

Step 6: Integrate each term

Integrate each of the terms separately: \( \int \frac{1/2}{u-1} \, du = \frac{1}{2} \ln |u-1| + C_1 \) and \( \int \frac{1/2}{u+1} \, du = \frac{1}{2} \ln |u+1| + C_2 \).

Step 7: Combine results and back-substitute

Combine the integrated results: \( \frac{1}{2} \ln |u-1| - \frac{1}{2} \ln |u+1| + C \). Simplify using the property of logarithms: \( \frac{1}{2} (\ln |u-1| - \ln |u+1|) = \frac{1}{2} \ln \left| \frac{u-1}{u+1} \right| \). Substitute back \( u = \exp(x) \) to get the final result: \( \frac{1}{2} \ln \left| \frac{\exp(x) - 1}{\exp(x) + 1} \right| + C \).

Key Concepts

Substitution Method

The substitution method is a mathematical technique often used to simplify complex integrals. It involves replacing a difficult expression with a single variable to make the integral easier to solve. In our given exercise, this technique is pivotal. Starting with an integral involving the exponential function \[\int \frac{\exp (x)}{\exp (2 x)-1} \, dx,\]we make a substitution to simplify the problem. By setting \( u = \exp(x) \), we reduce a potentially complex expression into a more manageable form. This substitution leads to \( du = \exp(x) \, dx \), allowing us to replace \( dx \) with \( \frac{du}{u} \). Once the substitution is made, the expression becomes significantly easier to deal with:\[\int \frac{du}{u^2 - 1}.\] Using substitution in integration involves:

• Identifying a part of the integrand that can be replaced with a single variable.
• Deriving a simple expression for \( dx \).
• Simplifying the integral in terms of the new variable.

Through this process, integrals that initially seem daunting become approachable.

Integration Techniques

Integration techniques are essential tools in calculus, used to find the antiderivatives of complex functions. Once we have simplified an integral using substitution, as we did by changing variables, further integration methods are required. In this exercise, partial fraction decomposition is used after substitution to break down the integral into simpler terms.Think of partial fractions as a way to split a complicated rational expression into a sum of simpler ones. For the expression \[\frac{1}{u^2 - 1},\]we observed that \( u^2 - 1 \) can be factored into \( (u-1)(u+1) \). Partial fraction decomposition lets us write it as:\[\frac{A}{u-1} + \frac{B}{u+1},\]where \( A \) and \( B \) are constants that we find by solving a system of equations. This method allows us to deal with more straightforward functions independently. Then, each part can be integrated separately:

• Integers of natural logarithmic forms like \( \ln |u-1| \).
• Handling each fraction as an independent problem.

With these techniques, even the most complex integrals become manageable, allowing precise calculation of integrals that initially appear unsolvable.

Exponential Functions

Exponential functions form a crucial part of mathematical studies, particularly when dealing with growth models and calculus. They have the form \( \exp(x) \) or \( e^x \), where \( e \) is a constant approximately equal to 2.71828. They are known for their rapid increase and have unique properties that differentiate them from other functions.In integration, exponential functions often appear, as in our original problem:\[\int \frac{\exp (x)}{\exp (2 x)-1} \, dx.\]Dealing with exponential functions requires recognizing patterns, such as how exponents behave under operations, e.g., \( \exp(2x) = (\exp(x))^2 \). This realization leads to our substitution \( u = \exp(x) \), simplifying our integral significantly.Key characteristics of exponential functions include:

• Continuous growth and always positive values.
• A derivative that is proportional to the function itself, meaning \( \frac{d}{dx}[e^x] = e^x \).
• Ability to model real-world phenomena like population growth or radioactive decay.

Mastering these functions can simplify the analysis of integrals and enable the application of more efficient solving strategies.