Problem 54 - Calculus Single Variable

Question

Make a substitution before applying the method of partial fractions to calculate the given integral. $$ \int \frac{2^{x+2}}{4^{x}-4} d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{4} \ln \left| \frac{2^x - 2}{2^x + 2} \right| + C \).

Step 1: Simplify the Expression

Recognize that \(4^x\) can be expressed as \((2^x)^2\), making the integral \( \int \frac{2^{x+2}}{(2^x)^2 - 4} \, dx \). Next, simplify the numerator: \(2^{x+2} = 4 \cdot 2^x\). Substitute into the integral to get \( \int \frac{4 \cdot 2^x}{(2^x)^2 - 4} \, dx \).

Step 2: Make a Substitution

Let \( u = 2^x \). Then \( du = 2^x \ln(2)\, dx \), or \( dx = \frac{du}{u \ln(2)} \). Substitute these into the integral: \( \int \frac{4u}{u^2 - 4} \cdot \frac{du}{u \ln(2)} = \frac{4}{\ln(2)} \int \frac{1}{u^2 - 4} \, du \).

Step 3: Factor the Denominator

The expression under the integral becomes \( \int \frac{1}{(u-2)(u+2)} \, du \). This denominator can be decomposed using partial fractions.

Step 4: Set Up Partial Fraction Decomposition

Assume \( \frac{1}{(u-2)(u+2)} = \frac{A}{u-2} + \frac{B}{u+2} \). Clear the fractions to get \( 1 = A(u+2) + B(u-2) \).

Step 5: Solve for Coefficients

Expand to get \( 1 = (A + B)u + (2A - 2B) \). By comparing coefficients, solve for \( A + B = 0 \) and \( 2A - 2B = 1 \). Solving these gives \( A = \frac{1}{4} \) and \( B = -\frac{1}{4} \).

Step 6: Integrate Using Partial Fractions

Substitute back into the integral to get:\[ \int \left( \frac{1/4}{u-2} - \frac{1/4}{u+2} \right) \, du = \frac{1}{4} \int \frac{1}{u-2} \, du - \frac{1}{4} \int \frac{1}{u+2} \, du. \]Integrate each term:\[ \frac{1}{4} \ln|u-2| - \frac{1}{4} \ln|u+2| + C. \]

Step 7: Substitute Back for \( u \)

Replace \( u \) with \( 2^x \) to convert back to the original variable:\[ \frac{1}{4} \ln|2^x - 2| - \frac{1}{4} \ln|2^x + 2| + C \].

Key Concepts

partial fraction decomposition

Partial fraction decomposition is a technique used to simplify complex rational expressions into simpler ones that are easier to integrate. When faced with a fraction where the denominator can be factored into linear or irreducible quadratic factors, partial fraction decomposition can be employed to break it down further.
Here's how it works:

• Express the original fraction as a sum of several fractions with simpler denominators.
• The numerators of these simpler fractions, typically constants or linear expressions, are then solved for to make the decomposition align with the original expression.
• This method is particularly useful because integrating these simpler terms separately is often more straightforward.

When using this technique, the goal is to obtain several smaller fractions, each with a denominator that consists of one of the original factors of the complex denominator. For example, in the exercise given, the factorization of the denominator leads to setting up equations to find the coefficients "A" and "B". Once found, the integration process becomes significantly simplified, allowing for straightforward application of basic integration techniques.

substitution method

The substitution method is a common strategy in integration that simplifies an integral using a change of variables. It aims to transform a difficult integral into a more familiar and manageable form.
Here's a simple breakdown of the steps involved:

• Identify a part of the integral that can be substituted by a new variable, making the integral easier to solve.
• Express the differential (in this case, "dx") in terms of the new variable "du".
• Substitute both the expression and the differential into the integral, then proceed to integrate with respect to the new variable "u".

After integration using the simpler expression, you must substitute back the original variable to complete the problem. In the exercise provided, letting 'u = 2^x' was a clever choice as it transformed the complex expression into one we could handle more easily. This strategic change laid the groundwork for simpler integration through partial fraction decomposition.

exponential functions

Exponential functions are mathematical functions of the form \(f(x) = a^x\), where \(a\) is a positive constant. They grow or decay rapidly based on the value of the base \(a\).
Key properties of exponential functions include:

• They are continuous and smooth.
• Over a certain domain, they are increasing if \(a > 1\) and decreasing if \(0 < a < 1\).
• The derivative of an exponential function joins beautifully with natural logarithms, as the derivative of \(e^x\) is \(e^x\).

In many integrative processes, transforming other exponential bases into powers of \(e\), the natural logarithmic base, can simplify differentiation and integration. In this particular exercise, expressing \(4^x\) as \((2^x)^2\) and working through that transformation were vital steps in attending to the integral efficiently.

logarithmic integration

Logarithmic integration is a special case of integration that often arises when you're working with rational functions or multiplicative forms of \(x\) related expressions. This technique directly benefits from derivatives and integrals of the natural logarithmic function.
The logarithmic rule for integration is:

• For an integral looking like \(\int \frac{1}{u} \, du\), the solution is \(\ln|u| + C\), where \(C\) is the constant of integration.
• When encountering a division that involves logarithms in the result, it is helpful to recognize the need for the absolute value inside the logarithm.

In the given exercise, once the integral is broken into partial fractions, each term takes the familiar form of \(\int \frac{1}{u} \, du\). Here, the technique easily leads us to obtain the natural logarithm expressions inside the final solution. After evaluating each term separately, the original variables are substituted back in, rendering a neat and tidy logarithmic integration answer. This illustrates the power of recognizing usable patterns and integrating accordingly.