Question

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

Short answer

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

Step-by-step solution

Substitution

We start by performing substitution to simplify the integral. To manage the exponential inside the integral, we can use a substitution where:Let \( u = e^{x} \), then \( du = e^{x} \, dx \) or equivalently \( dx = \frac{du}{u} \). Substituting these into the integral, we get:\[ \int \frac{1}{1 + e^{x}} \, dx = \int \frac{1}{1 + u} \cdot \frac{du}{u} = \int \frac{du}{u(1+u)} \].This transforms the integral into a rational function of \( u \).

Partial Fraction Decomposition

Next, decompose the rational function \( \frac{1}{u(1+u)} \) into partial fractions. We assume a decomposition of the form:\[ \frac{1}{u(1+u)} = \frac{A}{u} + \frac{B}{1+u} \].To find \( A \) and \( B \), multiply both sides by \( u(1+u) \) to clear the denominators:\[ 1 = A(1+u) + Bu \].Expanding the right side gives:\[ 1 = A + Au + Bu \].Combine like terms:\[ 1 = A + (A + B)u \].From the equation, we equate coefficients for the constant and linear terms. Solving for A and B gives:For the constant term, \( A = 1 \).For the linear term, \( A + B = 0 \).Substituting \( A = 1 \) implies \( 1 + B = 0 \) or \( B = -1 \).

Integrating Each Partial Fraction

Now that we have our partial fractions \( \frac{1}{u} - \frac{1}{1+u} \), we can integrate each one separately. This gives:\[ \int \left( \frac{1}{u} - \frac{1}{1+u} \right) du = \int \frac{1}{u} \, du - \int \frac{1}{1+u} \, du \].The integrals are:\[ \ln |u| - \ln |1+u| + C \],where \( C \) is the constant of integration.

Back-Substitute and Simplify

Substitute back \( u = e^{x} \) to express the solution in terms of \( x \):\[ \ln |e^{x}| - \ln |1+e^{x}| + C \].Since \( \ln |e^{x}| = x \) because \( e^{x} > 0 \) for all \( x \), the expression simplifies to:\[ x - \ln(1 + e^{x}) + C \].

Key concepts

Rational Functions

Rational functions are a fundamental concept in calculus, particularly when dealing with integrals. A rational function is simply a ratio of two polynomials. In the context of integration, expressing an integrand as a rational function can make it easier to evaluate.
For example: If we have a complex integrand involving exponential terms, like the function in the exercise \( \int \frac{1}{1+e^{x}} \, dx \), a substitution such as \( u = e^{x} \) can transform it into a rational function of \( u \):

• Numerator: Typically of lower degree than the denominator. Here, it's \( 1 \).
• Denominator: Composed of the polynomial \( 1 + u \).

Transforming expressions into rational functions simplifies them, which is groundwork for the partial fraction decomposition process.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational functions into simpler, more manageable parts. This is especially useful for integrating rational functions where direct integration isn't straightforward.
Here's the basic idea of how it's done:

• Assume a form for the decomposition: This involves expressing the function \( \frac{1}{u(1+u)} \) as \( \frac{A}{u} + \frac{B}{1+u} \).
• Clear the denominators by multiplying through by \( u(1+u) \) to find values for \( A \) and \( B \).
• Solve for the coefficients \( A \) and \( B \) by equating the coefficients from both sides of the equation.

Once decomposed, each fraction can be easily integrated separately. In this specific case, \( A \) was found to be 1, and \( B \) to be -1, leading to the simpler integrals \( \int \frac{1}{u} \, du \) and \( \int -\frac{1}{1+u} \, du \).

Integral Evaluation

Once partial fraction decomposition is completed, evaluating the integral becomes more straightforward. Each simplified fraction can be integrated individually. The process in this exercise is as follows:

• Integrate \( \frac{1}{u} \), leading to \( \ln|u| \).
• Integrate \( -\frac{1}{1+u} \), resulting in \( -\ln|1+u| \).

These integrations are basic logarithmic integrals. After integrating, the next step is to substitute back any initial variables to express the solution in the original variable, \( x \) in this case.
Finally, substitute \( u = e^x \) back to get the original variable expression: \(\ln|e^x| - \ln|1+e^x| + C\). This simplifies further to \( x - \ln(1+e^x) + C \). Doing so yields the integral's evaluated form, which is more comprehensible and elegant.