Question

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

Short answer

\( \int \frac{e^{x} \, dx}{e^{2x} - e^{x}} = \ln \left| 1 - \frac{1}{e^x} \right| + C \).

Step-by-step solution

Simplify the Expression

Notice that the expression inside the integral is \( \frac{e^x}{e^{2x} - e^x} \). This can be rewritten as \( \frac{e^x}{e^x(e^x - 1)} \), which simplifies to \( \frac{1}{e^x - 1} \). So the integral becomes \( \int \frac{1}{e^x - 1} \, dx \).

Use Substitution

Let \( u = e^x - 1 \). Then the derivative \( \frac{du}{dx} = e^x \) or \( du = e^x \, dx \). Since \( e^x = u + 1 \), we rewrite the integral as \( \int \frac{1}{u} \cdot \frac{du}{u+1} \).

Decomposition into Partial Fractions

Rewrite the integrand \( \frac{1}{u(u+1)} \) in terms of partial fractions: \( \frac{1}{u(u+1)} = \frac{A}{u} + \frac{B}{u+1} \). Equate \( 1 = A(u+1) + Bu \) and solve for \( A \) and \( B \).

Solve for Partial Fraction Coefficients

By equating coefficients, we find:1. For the constant term: \( A = 1 \).2. For the \( u \) term: \( A + B = 0 \), which gives \( B = -1 \). Therefore, the decomposition is: \( \frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u+1} \).

Integrate the Partial Fractions

Now integrate each part separately:\[ \int \left( \frac{1}{u} - \frac{1}{u+1} \right) \, du = \int \frac{1}{u} \, du - \int \frac{1}{u+1} \, du \]This becomes:\[ \ln|u| - \ln|u+1| + C = \ln\left|\frac{u}{u+1}\right| + C \]

Substitute Back to Original Variable

Recall that \( u = e^x - 1 \). Substitute back to find the solution in terms of \( x \):\[ \ln \left| \frac{e^x - 1}{e^x} \right| + C = \ln \left| 1 - \frac{1}{e^x} \right| + C \]

Key concepts

Substitution Method

The substitution method is a powerful technique for solving integrals. It involves changing the variable of integration to simplify the integral into a more manageable form. In this exercise, we initially deal with the integral \( \int \frac{e^x \, dx}{e^{2x} - e^x} \).
By recognizing that the expression \( \frac{e^x}{e^{2x} - e^x} \) can be simplified to \( \frac{1}{e^x - 1} \), we make the substitution \( u = e^x - 1 \). This leads to a new differential \( du = e^x \, dx \).
The integral is transformed into \( \int \frac{1}{u} \cdot \frac{du}{u+1} \), significantly simplifying the problem. Substitution helps us in this exercise to turn the exponential integral into a rational function form. This makes the remaining steps, like applying partial fractions, easier to tackle.

Partial Fractions

Partial fractions is a technique used to decompose rational functions into simpler fractions, making integration easier. It allows for the breakdown of a complicated rational function into a sum of simpler fractions.
In this exercise, after substitution, we reach the integral \( \int \frac{1}{u(u+1)} \, du \).
To integrate, we break it down using the partial fractions method:

• Write \( \frac{1}{u(u+1)} \) as \( \frac{A}{u} + \frac{B}{u+1} \).
• Equate and solve for constants A and B using: \(1 = A(u+1) + Bu\).

Simplifying, we derive \( A = 1 \) and \( B = -1 \).
These coefficients allow us to rewrite the integral as \( \int \frac{1}{u} \, du - \int \frac{1}{u+1} \, du \).
It turns a complex process into simpler terms, making integration doable.

Rational Functions

Rational functions are ratios of polynomials, often denoting a fraction where both the numerator and the denominator are polynomials.
In calculus, integrals with rational functions may require methods like partial fractions for evaluation. In this exercise, after employing substitution with \( u = e^x - 1 \), the integral simplifies to \( \int \frac{1}{u(u+1)} \, du \).
This expression represents a rational function since it is the fraction of two polynomial expressions.
By utilizing the partial fractions technique, this rational function is decomposed into simpler fractions. Thus, integration becomes straightforward by breaking it down into integrals of \( \frac{1}{u} \) and \( \frac{1}{u+1} \). Such transformation is key in calculus to solve complex integrals involving polynomial fractions.

Exponential Functions

Exponential functions are functions where the variable appears in the exponent and are of the form \( e^x \). They are particularly challenging to integrate when combined in complex expressions.
The given integral initially contains an exponential function, \( \frac{e^x}{e^{2x} - e^x} \), requiring simplification. Identifying the exponential elements aids in substituting to a rational form, notably with \( u = e^x - 1 \).
Following substitution, what started as exponential transforms into a rational function, easing integration using partial fractions. Exponential functions, thus, require careful handling through smart substitution and conversion techniques to simplify integration.
Understanding the behavior of exponential functions and their transformations is crucial for appropriately simplifying and solving integrals where they are involved.