Problem 71 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The cover-up method was used to find the constants A and B.

Step 1: Substitute a new variable

Let's substitute a new variable \(u\) for \(e^x\). We have: $$u = e^x$$ Differentiating both sides with respect to \(x\) we obtain: $$\frac{du}{dx} = e^x$$ Now, we can substitute the variable \(u\) into the integral: $$\int \frac{u}{(u-1)(u+2)} d u$$ Note that we have changed the differential notation from \(dx\) to \(du\).

Step 2: Perform partial fraction decomposition

To perform partial fractions decomposition, we rewrite the fraction \(\frac{u}{(u-1)(u+2)}\) as the sum of two simpler fractions \(\frac{A}{(u-1)}\) and \(\frac{B}{(u+2)}\): $$\frac{u}{(u-1)(u+2)} = \frac{A}{(u-1)} + \frac{B}{(u+2)}$$ Multiply both sides by \((u-1)(u+2)\) to clear denominators, we get: $$u = A(u+2) + B(u-1)$$ Now let's find the constants A and B.

Step 3: Find A and B using the cover-up method

Using the cover-up method, we set \(u = 1\): $$1 = A(1+2)$$ Solving for \(A\) gives us \(A = \frac{1}{3}\). Next, we set \(u = -2\): $$-2 = B(-2-1)$$ Solving for \(B\) gives us \(B = \frac{2}{3}\). So, the fraction becomes: $$\frac{\frac{1}{3}}{(u-1)} + \frac{\frac{2}{3}}{(u+2)}$$

Step 4: Integrate the simplified expression

Now let's integrate the two fractions: $$\int \frac{1/3}{(u-1)} du + \int \frac{2/3}{(u+2)} du$$ Integration of these fractions results in: $$\frac{1}{3} \ln |u-1| + \frac{2}{3} \ln |u+2| + C$$

Step 5: Substitute back in for x

Finally, substitute \(u = e^x\) back into the expression: $$\frac{1}{3} \ln |e^x-1| + \frac{2}{3} \ln |e^x+2| + C$$ This is the final expression of the integral.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex fractions into simpler ones, making integration easier. When you have a rational function where the numerator's degree is less than the denominator's, you can express it as a sum of fractions with simpler denominators.

Here's how it works:

• First, identify the linear factors in the denominator. For example, in our integral, the expression \((u-1)(u+2)\) is already factored into linear terms.
• Next, express the fraction \(\frac{u}{(u-1)(u+2)}\) as a sum of fractions: \(\frac{A}{(u-1)} + \frac{B}{(u+2)}\).
• Multiply through by the common denominator to eliminate fractions: \(u = A(u+2) + B(u-1)\).
• To find the constants \(A\) and \(B\), use methods like the cover-up method, which simplifies the process by choosing values of \(u\) that zero out the other term.

This process allows you to write complex fractions in a form that's straightforward to integrate, reflecting the power of partial fraction decomposition in solving integrals.

Change of Variables

The change of variables, or substitution method, is a powerful technique in integral calculus. It simplifies an integral by transforming it into a new variable, reducing complexity. For the integral \(\int \frac{e^{x}}{(e^{x}-1)(e^{x}+2)} \ dx\), we use substitution to make the math manageable.

Here's how we apply it:

• Choose a substitution that simplifies the expression. Here, we let \(u = e^x\), which simplifies the integral substantially.
• Differentiate the substitution with respect to \(x\), so \(\frac{du}{dx} = e^x\). Thus, \(dx\) is replaced by \(\frac{du}{e^x}\), simplifying to \(du\).
• Rewrite the integral in terms of \(u\), transforming \(\int \frac{e^{x}}{(e^{x}-1)(e^{x}+2)} \ dx \) into \(\int \frac{u}{(u-1)(u+2)} \ du\).

This substitution makes the integral easier to handle, setting the stage for applying further techniques like partial fraction decomposition.

Integration Techniques

Once you've simplified the integral using techniques such as substitution and partial fraction decomposition, integrating becomes straightforward. In our example, once we split the fraction, we deal with simpler fractions that are easier to integrate.

Let's explore the steps to integrate:

• After decomposing the fraction into \(\frac{\frac{1}{3}}{(u-1)} + \frac{\frac{2}{3}}{(u+2)}\), focus on integrating each part separately.
• Integrals of the form \(\int \frac{1}{ax+b} \ du\) result in \(\ln|ax+b|\). For example, \(\int \frac{1/3}{u-1} \ du = \frac{1}{3}\ln|u-1|\).
• Apply the same logic to the second term: \(\int \frac{2/3}{u+2} \ du = \frac{2}{3}\ln|u+2|\).

Once integrated, substitute back \(u = e^x\) to find the original integral in terms of \(x\).
This sequence simplifies complex integrals and showcases the versatility of integration techniques in calculus.