Problem 53 - College Algebra and Calculus: An Applied Approach

Question

Use any basic integration formula or formulas to find the indefinite integral. State which integration formula(s) you used to find the integral. $$ \int \frac{1+e^{-x}}{1+x e^{-x}} d x $$

Step-by-Step Solution

Answer

The indefinite integral of the given function is \( -x - ln|e^{-x}| + C \)

Step 1: Identify the form

In given integral \( \int \frac{1+e^{-x}}{1+x e^{-x}} dx \), it can be observed that the denominator is the derivative of the numerator. That concludes that a simple substitution method can solve the integral.

Step 2: Perform Substitution

Make the substitution \( u = 1 + e^{-x} \). Then the derivative of \( u \) with respect to \( x \) is \( du/dx = -e^{-x} \). Rearranging, \( du = -e^{-x} dx \), which implies \( dx = -du/e^{-x} \). Substituting the values of \( u \) and \( dx \) into the integral, it translates to \( \int \frac{u}{u-e^{-x}}(-du/e^{-x}) \).

Step 3: Simplify the Expression

Next, we simplify the integral. The \( e^{-x} \) in the denominator and numerator cancel out leaving us \( -\int \frac{u}{u-1} du \).

Step 4: Divide into simpler fractions

To further simplify, perform long division on the fraction which results in \( -\int (1+\frac{1}{u-1}) du \).

Step 5: Integration

Integrating term by term, we have \( -x - ln|u - 1| + C \).

Step 6: Back-substitute the value of u

Finally, substitute \( u = 1 + e^{-x} \) back into the equation, yielding \( -x - ln|e^{-x}| + C \) as the final answer.

Key Concepts

Integration by Substitution

Integration by substitution, also known as u-substitution, is a technique used in integral calculus that simplifies finding the antiderivative of complex functions. It is particularly helpful when dealing with integrals where the integrand is a composite function—essentially a function within another function.
By identifying a part of the integrand that can be considered as a separate function, say 'u', we can rewrite the integral in terms of u, which often results in a simpler integral that is easier to solve. To perform substitution effectively, one must decide what to substitute, find the derivative of that substitution to replace 'dx', and ensure that the integrand is entirely in terms of the new variable 'u' after substitution.
For example, in the exercise at hand, by substituting the more complex part of the fraction with 'u', the integral becomes much simpler to integrate. The derivative of the chosen 'u' also appears in the original integral, which is a clear indicator that the substitution is the correct approach. After substitution, the integration process involves simpler expressions that are more manageable to integrate, a typical goal when applying the substitution method.

Basic Integration Formulas

The foundation of calculating indefinite integrals is understanding basic integration formulas. These formulas provide the antiderivatives of elementary functions, such as power functions, exponential functions, and trigonometric functions. They are essential tools in any mathematician's toolkit and are especially useful in solving integrations that do not require complex techniques.

• For power functions of the form \(x^n\), the basic integration formula is \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), where 'n' is a real number different from -1 and 'C' represents the constant of integration.
• For the exponential function \(e^x\), the integral is \(\int e^x dx = e^x + C\).

Applying these fundamental formulas allows one to swiftly compute integrals without the need for elaborate techniques. In the exercise, recognizing that the derivative of \(e^{-x}\) relates closely to the integral's denominator enabled the use of these formulas after applying substitution. This is a clear example of how basic integration formulas are used in tandem with other methods to solve integrals.

Integrating Algebraic Fractions

When we encounter algebraic fractions in integrals, they may initially appear daunting to integrate. However, by breaking them down into simpler parts or by using various techniques like substitution, integrating these fractions becomes much more manageable.
An algebraic fraction typically involves a numerator and a denominator that are polynomials. One common approach to integrate such expressions is to perform polynomial division if the degree of the numerator is higher than the denominator. An alternative is to decompose the fraction into partial fractions if the denominator can be factorized. These strategies transform the original complex fraction into a sum of simpler fractions that are much easier to integrate.
In the given exercise, after substitution, the integrand changed from a more complex fraction to a simpler one that could be divided into two terms: \(1\) and \(\frac{1}{u-1}\). These terms are straightforward to integrate. The first term integrates to 'x', and the second term, being the derivative of a natural logarithm function, integrates to \(ln|u-1|\), thus demonstrating a practical application of integrating algebraic fractions in integral calculus.