Question

Preliminary steps The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals. $$\int \frac{e^{x}}{\left(e^{x}-1\right)\left(e^{x}+2\right)} d x$$

Short answer

Question: Evaluate the integral: $$\int \frac{e^{x}}{\left(e^{x}-1\right)\left(e^{x}+2\right)} dx$$ Answer: The integral evaluates to: $$\int \frac{e^{x}}{\left(e^{x}-1\right)\left(e^{x}+2\right)} dx = \frac{1}{3}\ln\left|\frac{e^x-1}{e^x+2}\right| + C$$

Step-by-step solution

Change of Variables

Let's perform a change of variables, letting \(u = e^x\). Then, \(\frac{du}{dx} = e^x\) or \(dx = \frac{du}{u}\). The integral then becomes: $$\int \frac{u}{(u-1)(u+2)} \frac{du}{u}$$ Cancelling the u's, we get: $$\int \frac{1}{(u-1)(u+2)} du$$

Partial Fraction Decomposition

Now, we use the method of partial fractions to decompose the integrand. We write: $$\frac{1}{(u-1)(u+2)} = \frac{A}{u-1} + \frac{B}{u+2}$$ To determine the constants A and B, we multiply both sides by \((u-1)(u+2)\) to eliminate the denominators: $$1 = A(u+2) + B(u-1)$$

Solve for A and B

We can now solve for A and B by solving the system of linear equations. Let's set \(u-1\) equal to zero, i.e., \(u=1\): $$1 = A(1+2) + B(1-1) \Rightarrow A = \frac{1}{3}$$ Now, let's set \(u+2\) equal to zero, i.e., \(u=-2\): $$1 = A(-2+2) + B(-2-1) \Rightarrow B = -\frac{1}{3}$$ Our partial fractions decomposition is then: $$\frac{1}{(u-1)(u+2)} = \frac{1/3}{u-1} - \frac{1/3}{u+2}$$

Integrate

Now, we can integrate each term with respect to u: $$\int \frac{1}{3(u-1)} du - \int \frac{1}{3(u+2)} du$$ Using the integral property: $$\frac{1}{3}\int \frac{1}{u-1} du - \frac{1}{3}\int \frac{1}{u+2} du$$ Integrating the terms, we have: $$\frac{1}{3}\ln|u-1| - \frac{1}{3}\ln|u+2| + C$$

Reverse Substitution

We should now reverse the substitution by replacing \(u\) with \(e^x\): $$\frac{1}{3}\ln|e^x-1| - \frac{1}{3}\ln|e^x+2| + C$$ Now we have evaluated the integral: $$\int \frac{e^{x}}{\left(e^{x}-1\right)\left(e^{x}+2\right)} dx = \frac{1}{3}\ln\left|\frac{e^x-1}{e^x+2}\right| + C$$

Key concepts

Change of Variables

The concept of change of variables is a strategic move in calculus to simplify complex integrals. Instead of tackling a tough integral head-on, we cleverly substitute a part of the integral with a new variable, often denoted as 'u'. This makes the integral look different—hopefully easier to work with.

Here's how it works: We identify a function within the integral that we can replace with 'u', which in turn simplifies the integral. We then differentiate this function with respect to 'x' to find 'du', and from there, we can express 'dx' in terms of 'du'. The magic happens when our integral, with its new variable, looks much simpler and more approachable for further steps, like partial fraction decomposition or direct integration. This little sleight of hand not only cleans up the math but often reveals easier paths to the solution.

Partial Fraction Decomposition

Partial fraction decomposition is a technique that takes a complex rational expression and breaks it down into simpler fractions that are easier to integrate. Imagine you're trying to eat a big slice of pizza. It's too big to eat in one bite, so what do you do? You slice it up into smaller, manageable pieces. That's exactly what partial fraction decomposition does with integrals.

To perform this culinary magic trick with integrals, we look for constants (like 'A' and 'B' in the solution) that, when plugged into our simpler fractions, reconstruct the original complex fraction. It requires a bit of algebra, setting up equations that hold true for all values of 'u', and finally solving these equations to find the constants. Once we have these constants, each piece of our now-fractured integral can be integrated separately, making the whole process much more digestible.

Integration Techniques

There are various tools at our disposal within the toolbox of integration techniques. These techniques include substitution (change of variables), partial fraction decomposition, integration by parts, and trigonometric substitution, among others. Each serves a unique purpose, and recognizing which tool to use is part art, part science.

In our exercise, we're using a one-two punch of change of variables followed by partial fraction decomposition—each technique complements the other to break down our integral into more tractable parts. This orchestrated approach is a cornerstone of integral calculus; knowing how and when to apply these techniques can transform a seemingly intractable problem into a series of solvable puzzles.

Exponential Functions

Exponential functions involve numbers raised to a variable power, like 'e^x'. They are as essential to calculus as bread is to a sandwich. These functions have unique properties, such as the rate of increase being proportional to the value of the function itself—as exemplified by natural growth processes or compound interest.

In our case, 'e^x' is part of both the numerator and the denominator, giving us a key insight to perform an effective change of variables. The beauty of exponential functions extends to their derivatives and integrals, as they maintain their exponential form. This consistent behavior is a valuable trait that allows for simplification during integration, particularly when the exponent itself is a straightforward expression, as is 'x' in this exercise. Because they play so nicely in the realm of differentiation and integration, exponential functions are a central theme in the study of calculus.