Question

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. \(\int \frac{1}{2+e^{-x}} d x\)

Short answer

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

Step-by-step solution

Choose a substitution

To simplify the integral \( \int \frac{1}{2+e^{-x}} \, dx \), we start by making a substitution. Let \( u = e^{-x} \). Consequently, the derivative \( du = -e^{-x} \, dx \) or \( dx = \frac{-du}{u} \). Substitute these into the integral.

Substitute into the integral

Substituting \( u = e^{-x} \) and \( dx = \frac{-du}{u} \) transforms the integral: \[ \int \frac{1}{2+u} \, \left(\frac{-du}{u}\right) = -\int \frac{1}{u(2+u)} \, du \]. This is now an integral of a rational function.

Decompose into partial fractions

To integrate the rational function \( \frac{1}{u(2+u)} \), we decompose it into partial fractions. Assume \( \frac{1}{u(2+u)} = \frac{A}{u} + \frac{B}{2+u} \). Multiply both sides by the denominator \( u(2+u) \) to get \( 1 = A(2+u) + Bu \).

Solve for the coefficients

Substitute values to solve for \( A \) and \( B \). Let \( u = 0 \), which gives \( A = \frac{1}{2} \). Let \( u = -2 \), which gives \( B = \frac{1}{2} \). Thus, \( \frac{1}{u(2+u)} = \frac{1/2}{u} + \frac{1/2}{2+u} \).

Integrate the partial fractions

Now integrate each part separately: \[ -\int \left( \frac{1/2}{u} + \frac{1/2}{2+u} \right) du = -\frac{1}{2} \int \frac{1}{u} \, du - \frac{1}{2} \int \frac{1}{2+u} \, du \].

Compute the integrals

These integrals are basic logarithmic integrals: \( -\frac{1}{2} \ln|u| - \frac{1}{2} \ln|2+u| \). Combine constants: \[ = -\frac{1}{2}( \ln |u| + \ln |2+u|) = -\frac{1}{2} \ln (|u(2+u)|) + C \].

Substitute back the original variable

Recall \( u = e^{-x} \). Substitute back: \( -\frac{1}{2} \ln |e^{-x} (2 + e^{-x})| = -\frac{1}{2} ( -x + \ln |2 + e^{-x}| ) + C \). After simplification, \( \frac{x}{2} - \frac{1}{2} \ln |2 + e^{-x}| + C \) is the final solution.

Key concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify complex integrals. The idea is to change the variable of integration, transforming the integral into a simpler form that is easier to evaluate. For the given integral, \( \int \frac{1}{2+e^{-x}} \, dx \), we use the substitution \( u = e^{-x} \). This choice is strategic because it simplifies the exponential term in the denominator.
You also find the differential \( du = -e^{-x} \, dx \), which leads to \( dx = \frac{-du}{u} \). This variable change yields a new integral: \( -\int \frac{1}{u(2+u)} \, du \).
This substitute transforms the original integral into an expression featuring a rational function, preparing it for further techniques like partial fraction decomposition:

• Choose a substitution that simplifies the expression.
• Determine the new differential to replace \( dx \).
• Substitute the new variables and differentials into the original integral.

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used to integrate rational functions, expressions of the form \( \frac{P(x)}{Q(x)} \). This technique involves breaking down a complex rational expression into a sum of simpler fractions. In our problem, we decompose \( \frac{1}{u(2+u)} \) into simpler fractions.
Assume \( \frac{1}{u(2+u)} = \frac{A}{u} + \frac{B}{2+u} \). To find \( A \) and \( B \), multiply both sides by the denominator \( u(2+u) \), getting \( 1 = A(2+u) + Bu \).
Next, solve for the values of \( A \) and \( B \) by substituting convenient values for \( u \), such as \( u = 0 \) and \( u = -2 \), yielding \( A = \frac{1}{2} \) and \( B = \frac{1}{2} \).

• Express the rational function as a sum of simpler fractions.
• Solve for the coefficients by equating terms with the same powers or specific substitution.
• Use these decomposed fractions for integration.

Rational Functions

Rational functions are ratios of two polynomials, where the degree of the numerator is less than or equal to the degree of the denominator. In this exercise, rational functions appear after substitution transforms the integral into \( \int \frac{1}{u(2+u)} \, du \).
The decomposition process allows us to work with simpler rational functions, such as \( \frac{1/2}{u} + \frac{1/2}{2+u} \). These simpler fractions are much easier to integrate. Rational functions frequently require methods like substitution or partial fraction decomposition for integration.

• Simplify complex fractions into manageable parts.
• Use algebraic methods to work through function transformations.
• Apply integration techniques to each part separately for solution finding.

Logarithmic Integration

Logarithmic integration is a technique used when integrating fractions where the denominator is a simple linear term. Once we decompose \( \frac{1}{u(2+u)} \) into partial fractions, each fraction is of a form that can be integrated using basic rules of integration involving logarithms.
Integrate each part separately: \( -\frac{1}{2} \int \frac{1}{u} \, du \) and \( -\frac{1}{2} \int \frac{1}{2+u} \, du \). These integrals result in \(-\frac{1}{2} \ln|u|\) and \(-\frac{1}{2} \ln|2+u|\), respectively.
Finally, these results are combined to form the solution \( -\frac{1}{2} (\ln |u| + \ln |2+u|) \), simplified to \( -\frac{1}{2} \ln |u(2+u)| + C \).
When transformed back to the original variables, the final expression involves logarithms, showing how logarithmic integration simplifies and completes the integration process for these kinds of rational functions.

• Simplify fractions into forms suitable for logarithmic integration.
• Apply basic logarithmic integration rules to each term.
• Re-substitute original variables at the end of integration.