Question

The following integrals require a preliminary step such as long division or a change of variables before using 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 $\int \frac{e^{x}}{\left(e^{x}-1\right)\left(e^{x}+2\right)} d x$ Answer: $\frac{1}{3}\ln|e^x-1| - \frac{1}{3}\ln|e^x+2| + C$

Step-by-step solution

Substitution

Let us substitute \(u = e^x\), so we have \(du = e^x\, dx\). Now we can rewrite the integral as: $$\int \frac{1}{\left(u-1\right)\left(u+2\right)} \, du$$

Partial Fractions

Using the method of partial fractions, we will decompose the fraction into simpler fractions: $$\frac{1}{\left(u-1\right)\left(u+2\right)} = \frac{A}{u-1} + \frac{B}{u+2}$$ To find the constants A and B, we will clear the denominators: $$1 = A(u+2) + B(u-1)$$ Now we can find A and B by choosing values for u that will make the other term zero. For A, let \(u = 1\): $$1 = 3A \Rightarrow A = \frac{1}{3}$$ For B, let \(u = -2\): $$1 = -3B \Rightarrow B = -\frac{1}{3}$$ Now we can rewrite the integral with the partial fraction decomposition: $$\int \left(\frac{\frac{1}{3}}{u-1} - \frac{\frac{1}{3}}{u+2}\right) \, du$$

Integrate

Now, we can integrate the two simpler fractions: $$\int \frac{1/3}{u-1} \, du - \int \frac{1/3}{u+2} \, du$$ The resulting integral is: $$\frac{1}{3}\ln|u-1| - \frac{1}{3}\ln|u+2| + C$$

Back Substitute

Finally, we'll back substitute the original variable, \(u = e^x\): $$\frac{1}{3}\ln|e^x-1| - \frac{1}{3}\ln|e^x+2| + C$$ This is the final solution for the given integral.

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify complex rational expressions, making them easier to integrate. The idea is to express the fraction as the sum of simpler fractions. This is particularly useful when dealing with fractions in the integrands of calculus problems.

Here's a straightforward way to approach partial fractions:

• Identify the form of the rational expression. Ensure the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division first.
• Factor the denominator completely. In the given example, we have the factors \(u-1\) and \(u+2\).
• Set up the partial fraction equation by writing a sum of fractions with unknown coefficients (e.g., \( \frac{A}{u-1} + \frac{B}{u+2} \)).
• Eliminate the denominators to solve for the coefficients by plugging in values for variables that simplify the equation.

By understanding and using this method, complex integrals can often be broken down into easier components, allowing for straightforward integration.

Substitution Method

The substitution method is a common integration technique that involves changing variables to simplify an integral. This process often involves substituting a part of the integrand with a new variable. The goal is to transform a difficult integral into a simpler form.

Here's how substitution works:

• Select a suitable substitution. For instance, in the given problem, we substitute \( u = e^x \).
• Differentiate the substitution equation to find the differential in terms of the new variable, which is \( du = e^x \, dx \).
• Rewrite the entire integral in terms of the new variable, replacing \( x \) and \( dx \) with \( u \) and \( du \).

This approach not only makes the integral more manageable but also allows for the use of additional techniques like partial fraction decomposition, as seen in this example.

Logarithmic Integration

Logarithmic integration involves integrating functions that produce a logarithmic function upon integration. This technique often applies when the integrand has the form of \( \frac{1}{x-a} \).

Key aspects of logarithmic integration:

• Recognize when a rational function can be expressed as a logarithm. For instance, \( \int \frac{1}{3(u-1)} \, du = \frac{1}{3}\ln|u-1| \).
• Apply the natural logarithm rule for integration: \( \int \frac{1}{u} \, du = \ln|u| + C \).
• Use integration by parts or substitution when the problem is more complex, eventually leading to a logarithmic form.

This technique was used in the solution by integrating the decomposed partial fractions into terms involving natural logarithms, simplifying the expression further.

Definite and Indefinite Integrals

Definite and indefinite integrals are two fundamental concepts in calculus. Understanding the distinction between them and their applications is crucial.

• An indefinite integral, such as the given problem, represents a family of functions and includes a constant of integration (denoted by \(C\)). It is typically expressed as \( \int f(x) \, dx = F(x) + C \, \) where \( F(x) \) is an antiderivative of \( f(x) \).
• On the other hand, a definite integral calculates the net area under a curve between two points. It is expressed with limits \( a \) and \( b \): \( \int_a^b f(x) \, dx \).
• The indefinite integral solution in this problem is \( \int \frac{1}{(e^x-1)(e^x+2)} \, dx = \frac{1}{3} \ln|e^x-1| - \frac{1}{3} \ln|e^x+2| + C \), showcasing the process of finding an antiderivative without evaluation over specific bounds.

Both types of integrals are instrumental in analyzing a variety of physical phenomena, from calculating area and volume to solving differential equations.