Problem 27 - Essential Calculus

Question

Use substitution to find the integral. $$ \int \frac{e^{x}}{\left(e^{x}-1\right)\left(e^{x}+4\right)} d x $$

Step-by-Step Solution

Answer

The solution to the integral is \( \frac{1}{3} \log{|e^x-1|} - \frac{1}{3} \log{|e^x+4|} + C \)

Step 1: Substitution

We start by substituting \( e^x = u \). Thus, \( dx = du/u \) as \( du = e^x \, dx \). Therefore, the integral becomes \( \int \frac{u}{(u-1)(u+4)} \, du \)

Step 2: Split into Partial Fractions

The integral now can be split into partial fractions. We would solve \(\frac{u}{(u-1)(u+4)} = \frac{A}{u-1} + \frac{B}{u+4}\) for the constants A and B. By solving this, we then find that A = 1/3 and B = -1/3. Thus, the integral becomes \( \int\frac{1}{3(u-1)} - \frac{1}{3(u+4)} \, du \)

Step 3: Integrating the Partial Fractions

We can now integrate each of the fractions separately: \( \frac{1}{3} \, \int \frac{1}{u-1} \, du - \frac{1}{3} \, \int \frac{1}{u+4} \, du \). Solving these gives \( \frac{1}{3} \log{|u-1|} - \frac{1}{3} \log{|u+4|} + C \) where C is the constant of integration.

Step 4: Back substitution

Substitute \( u = e^x \) back into the solution to get the final result in term of x: \( \frac{1}{3} \log{|e^x-1|} - \frac{1}{3} \log{|e^x+4|} + C \)

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a powerful algebraic technique used in calculus to break down rational expressions into simpler fractions that are easier to integrate. In our given integral, \[\int \frac{u}{(u-1)(u+4)} \, du\]we notice that the denominator can be factored into two linear terms, \((u-1)(u+4)\). By using partial fraction decomposition, we attempt to express this fraction as:

• \(\frac{A}{u-1} + \frac{B}{u+4}\)

The goal is to solve for the constants A and B, which essentially involves equating the original fraction to the sum of these simpler fractions and finding the values of A and B through algebraic methods. For this exercise, solving gives us A = 1/3 and B = -1/3, allowing us to rewrite the integrand in a form that can each be integrated easily.

Integration Constants

When we calculate indefinite integrals, we include a term known as the "constant of integration," typically represented by the letter C. This constant accounts for any vertical shift in the antiderivative since differentiation eliminates constants.

• In our solution: \(\frac{1}{3} \log{|u-1|} - \frac{1}{3} \log{|u+4|} + C\), the term C represents all potential vertical shifts.
• This means that for any given x, there exist infinitely many antiderivatives corresponding to different values of C.

This constant becomes particularly important when dealing with definite integrals, which compute specific values over an interval and therefore don't need C, as opposed to indefinite integrals that imply an entire family of functions.

Natural Logarithm Properties

Natural logarithms, denoted as \(\ln(x)\) or \(\log_e{x}\), have several properties that simplify integration and algebraic manipulations:

• \(\ln(ab) = \ln(a) + \ln(b)\)
• \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
• \(\ln(a^b) = b\ln(a)\)

In our integral solution, these properties help simplify the expression. For example, after integrating each term, the expression \(\frac{1}{3} \log{|u-1|} - \frac{1}{3} \log{|u+4|}\) could be combined using logarithmic properties to a single logarithm form. Understanding these properties can simplify many problems involving natural logs.

Definite and Indefinite Integrals

In calculus, integrals are classified as definite or indefinite:

• Indefinite integrals, such as \(\int f(x) \, dx\), provide a family of functions (antiderivatives) and include a constant of integration, C.
• In our exercise, the substitution and subsequent integration with a constant is an example of an indefinite integral.
• Definite integrals, like \(\int_{a}^{b} f(x) \, dx\), calculate the signed area under the curve f(x) from a to b and provide a numerical result. They do not include a constant of integration because they represent a specific value.

The step-by-step approach discussed focuses on finding indefinite integrals using substitution and partial fractions. Knowing when to apply definite versus indefinite integrals is key to solving calculus problems effectively.