Problem 55 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. $$\int \frac{e^{x}}{36-e^{2 x}} d x, x<\ln 6$$

Step-by-Step Solution

Answer

To find the indefinite integral of the given function, we followed these steps: 1. Applied substitution rule: \(u = e^x\), \(du = e^x dx\) 2. Factored the denominator: \(36 - u^2 = (6 - u)(6 + u)\) 3. Applied partial fraction decomposition: \(\frac{1}{(6 - u)(6 + u)} = \frac{1/12}{6 - u} + \frac{1/12}{6 + u}\) 4. Applied the antiderivative formula and found the antiderivative: \(-\frac{1}{12}\log|6 - u| + \frac{1}{12} \log |6 + u| + C\) 5. Applied back substitution: \(u = e^x\) The indefinite integral of the given function for \(x < \ln 6\) is: $$\int \frac{e^{x}}{36-e^{2 x}} dx = -\frac{1}{12}\log|6 - e^x| + \frac{1}{12} \log |6 + e^x| + C$$

Step 1: Apply substitution rule

First, we substitute \(u\) for \(e^x\), and denote its differential as \(du\). From this, we have: $$u = e^x \Rightarrow du = e^x dx$$ Now the integral becomes: $$\int \frac{du}{36-u^2}$$

Step 2: Factor the denominator

Notice that we can factor the denominator like this: $$36 - u^2 = (6 - u)(6 + u)$$ So the integral becomes: $$\int \frac{du}{(6 - u)(6 + u)}$$

Step 3: Apply partial fraction decomposition

Let's decompose the fraction into partial fractions. We look for constants A and B such that: $$\frac{1}{(6 - u)(6 + u)} = \frac{A}{6 - u} + \frac{B}{6 + u}$$ Now, to isolate A, we multiply both sides by \((6 - u)\) and evaluate \(u = 6\): $$A = \frac{1}{6 + 6} = \frac{1}{12}$$ Similarly, to isolate B, we multiply both sides by \((6 + u)\) and evaluate \(u = -6\): $$B = \frac{1}{6 - (-6)} = \frac{1}{12}$$ So the partial fraction decomposition is: $$\frac{1}{(6 - u)(6 + u)} = \frac{1/12}{6 - u} + \frac{1/12}{6 + u}$$ Now the integral becomes: $$\int \left(\frac{1/12}{6 - u} + \frac{1/12}{6 + u}\right) du$$

Step 4: Apply the antiderivative formula

Find the antiderivative for each term: $$\int \left(\frac{1/12}{6 - u} + \frac{1/12}{6 + u}\right) du = \frac{1}{12} \int \frac{1}{6 - u} du + \frac{1}{12} \int \frac{1}{6 + u} du$$ Now, apply the formula: $$\int \frac{1}{ax + b} dx = \frac{1}{a} \log |ax + b| + C$$ On the first integral, we have \(a = -1\), \(b = 6\), and on the second integral, \(a = 1\), \(b = 6\): $$\frac{1}{12} \int \frac{1}{6 - u} du = -\frac{1}{12}\log|6 - u| + C_1$$ $$\frac{1}{12} \int \frac{1}{6 + u} du = \frac{1}{12} \log |6 + u| + C_2$$ So, the antiderivative is: $$-\frac{1}{12}\log|6 - u| + \frac{1}{12} \log |6 + u| + C$$

Step 5: Apply the back substitution

Substitute back \(u = e^x\): $$-\frac{1}{12}\log|6 - e^x| + \frac{1}{12} \log |6 + e^x| + C$$ This is the indefinite integral of the given function for \(x < \ln 6\): $$\int \frac{e^{x}}{36-e^{2 x}} dx = -\frac{1}{12}\log|6 - e^x| + \frac{1}{12} \log |6 + e^x| + C$$

Key Concepts

substitution rule

The substitution rule is a powerful technique in calculus that simplifies complex integrals into more manageable forms. This involves selecting a substitution that introduces a new variable, usually denoted as \( u \). This substitution simplifies the algebraic expression, making the integration process more straightforward.

In the context of the given exercise, we started with the integral \( \int \frac{e^x}{36 - e^{2x}} \, dx \). By using the substitution \( u = e^x \), we transformed the integral into \( \int \frac{du}{36 - u^2} \).

• The substitution \( u = e^x \) replaces \( e^x \) with \( u \).
• The corresponding differential \( du = e^x \, dx \) simplifies \( dx \) in terms of \( du \).

This new form of the integral is much easier to tackle, illustrating how substitution helps unravel the original complex expression. The simplicity introduced here is key to making further calculations more efficient and clear.

partial fraction decomposition

Partial fraction decomposition is an algebraic technique used to break down rational expressions into simpler fractions. These simpler fractions are easier to work with, particularly when integrating. In our problem, the expression \( \int \frac{du}{(6 - u)(6 + u)} \) was incomplete without decomposition.

Here's how it's done:

• You express the fraction \( \frac{1}{(6-u)(6+u)} \) in the form of \( \frac{A}{6-u} + \frac{B}{6+u} \).
• By solving for constants \( A \) and \( B \), we isolate terms which allows integration term by term later on.
• We specifically solved \( A = \frac{1}{12} \) and \( B = \frac{1}{12} \).

The decomposition transforms the integrand into a sum of simpler fractions: \( \frac{1/12}{6-u} + \frac{1/12}{6+u} \). This now allows straightforward integration using basic logarithmic integration rules. It's a crucial step that often simplifies complex rational expressions.

antiderivative

Finding the antiderivative is often referred to as finding the indefinite integral. This involves determining a function whose derivative would yield the integrand. In the presented exercise, once partial decomposition was accomplished, the antiderivative was sought for each simpler term.

With the format \( \int \frac{1}{6-u} \, du \) and \( \int \frac{1}{6+u} \, du \), the antiderivatives individually become logarithmic functions as follows:

• For \( \int \frac{1}{6-u} \, du \), using the formula \( \int \frac{1}{ax + b} \, dx = \frac{1}{a} \log |ax + b| + C \), we derived \( -\frac{1}{12}\log|6-u| + C_1 \).
• For \( \int \frac{1}{6+u} \, du \), similarly, it became \( \frac{1}{12}\log|6+u| + C_2 \).

These logarithmic terms denote the antiderivative because their derivatives rearrange back to the original function. Therefore, the complete antiderivative becomes a sum of these two expressions, eventually leading to the solution when simplifying further.

integration by substitution

Integration by substitution is a logical extension of the substitution rule, specifically tailored to integrate functions that may initially appear too complicated. After employing substitution, you essentially reverse the process to fit the original variable back into the solution.

In our solution:

• After the antiderivatives were computed using \( u = e^x \), we substituted back \( u = e^x \) into \( -\frac{1}{12}\log|6-u| + \frac{1}{12}\log|6+u| + C \).
• This yields \( -\frac{1}{12}\log|6-e^x| + \frac{1}{12}\log|6+e^x| + C \).

By substituting \( u \) back to \( e^x \), one restores the variable in the context of the original function, allowing the result to represent the integration of the initial problem completely. Integration by substitution is a key step to ensure that the solution pertains to the original variable context, making the result much more relevant and usable.