Question

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

Short answer

The integral evaluates to \( x - 2\ln |1-e^x| + C \).

Step-by-step solution

Choose a Substitution

To simplify the integral, we'll use the substitution \( u = e^x \). This implies that \( du = e^x \, dx \), or equivalently, \( dx = \frac{du}{u} \). Substitute these into the integral.

Rewrite the Integral

With \( u = e^x \), the integral \( \int \frac{1+e^x}{1-e^x} \, dx \) becomes \( \int \frac{1+u}{1-u} \cdot \frac{du}{u} \). Simplify this to obtain \( \int \frac{1+u}{u(1-u)} \, du \).

Use Partial Fraction Decomposition

The integrand is now \( \frac{1+u}{u(1-u)} \). We need to express this as a sum of partial fractions: \( \frac{1+u}{u(1-u)} = \frac{A}{u} + \frac{B}{1-u} \). Multiply through by the denominator \( u(1-u) \) to obtain \( 1+u = A(1-u) + Bu \).

Solve for A and B

Expand and simplify the equation from the previous step: \( 1+u = A - Au + Bu \). Group like terms to obtain: \( 1+u = A + (B-A)u \). Equate coefficients to solve for A and B. This gives \( A = 1 \) and \( B-A = 1 \), so \( B = 2 \).

Integrate the Partial Fractions

Substitute \( A \) and \( B \) back into the partial fractions: \( \frac{1}{u} + \frac{2}{1-u} \). The integral becomes \( \int \left( \frac{1}{u} + \frac{2}{1-u} \right) \, du \). This separates into two integrals: \( \int \frac{1}{u} \, du + 2 \int \frac{1}{1-u} \, du \).

Evaluate the Integrals

Evaluate the integrals: \( \int \frac{1}{u} \, du = \ln |u| \) and \( \int \frac{1}{1-u} \, du = -\ln |1-u| \). So the original integral becomes \( \ln |u| - 2\ln |1-u| + C \), where \( C \) is the constant of integration.

Back Substitution

Since \( u = e^x \), substitute back to express the result in terms of \( x \): \( \ln |e^x| - 2\ln |1-e^x| + C \). Simplify using the properties of logarithms to obtain \( x - 2\ln |1-e^x| + C \).

Key concepts

Substitution Method

The substitution method is a helpful technique in integral calculus for simplifying integrals. By selecting a new variable, we can transform a complicated expression into a more manageable form. In the case of our exercise, the initial integral is given in terms of the variable \( x \). We choose a substitution, such as \( u = e^x \), to simplify the process. This means:

• The variable \( u \) is substituted for \( e^x \).
• The differential \( du \) relates to \( dx \). Specifically, we find that \( du = e^x \, dx \), which implies \( dx = \frac{du}{u} \).

By substituting these into the integral, we convert it to a form that is often easier to handle. The main idea is to simplify the integrand, making the next steps more straightforward.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify the integration of rational functions by breaking them down into simpler fractions. This is particularly useful in our exercise once the substitution method has been applied. Our integrand, after substitution, becomes a rational function \( \frac{1+u}{u(1-u)} \).

To apply partial fraction decomposition, we express the function as:\[ \frac{1+u}{u(1-u)} = \frac{A}{u} + \frac{B}{1-u} \]

• Multiply through by the common denominator, \( u(1-u) \), to eliminate the fractions and create an equation with known coefficients.
• Set up an equation based on the equality of polynomials: \( 1+u = A(1-u) + Bu \).

By expanding and grouping like terms, we determine the values of \( A \) and \( B \). Solve these coefficient equations simultaneously to get \( A = 1 \) and \( B = 2 \). This allows us to rewrite the original function into simpler components, making the integration process simpler.

Rational Functions

Rational functions are expressions made up of a ratio of two polynomials. These functions often appear in calculus, where their integration may require multiple methods. In our example, we encounter a rational function after applying the substitution method:\[ \frac{1+u}{u(1-u)} \]These functions are ideal candidates for partial fraction decomposition, as they can be expressed as sums of simpler fractions. This transformation is crucial since it allows us to implement basic integration techniques on the individual terms.

Once decomposed, each fraction can be integrated separately. For instance, in the simplified forms:

• The \( \frac{1}{u} \) part integrates into \( \ln|u| \).
• Similarly, the \( \frac{2}{1-u} \) integrates into \(-2\ln|1-u| \).

Through these steps, solving integrals involving rational functions becomes more approachable and clearer.