Problem 68 - Calculus Early Transcendentals

Question

The following integrals require a preliminary step such as long division or a change of variables before using the method of partial fractions. Evaluate these integrals. $$\int \frac{d x}{e^{x}+e^{2 x}}$$

Step-by-Step Solution

Answer

Based on the step by step solution above, determine the integral of the given function: $$\int \frac{d x}{e^{x}+e^{2 x}}$$ Solution: The integral of the given function is: $$\int \frac{d x}{e^{x}+e^{2 x}} = x - \ln (e^x + 1) + C$$

Step 1: Find a suitable substitution

We can simplify the expression inside the integral by substituting \(u = e^x\). This gives us \(\frac{1}{u^2+u}\).

Step 2: Find the derivative of the substitution

We have \(u = e^x\). Differentiating both sides with respect to \(x\), we get: $$\frac{d u}{d x} = e^x$$ Now, we need to substitute for \(dx\). Since we have \(\frac{d u}{d x} = e^x\), we can write \(d x = \frac{d u}{e^x}\).

Step 3: Rewrite the integral in terms of \(u\)

We substitute \(u = e^x\), and \(d x = \frac{d u}{e^x}\) in the integral. This gives us: $$\int \frac{d x}{e^{x}+e^{2 x}} = \int \frac{du}{u(u+1)}$$

Step 4: Perform partial fraction decomposition

We need to decompose the expression inside the integral, \(\frac{1}{u(u+1)}\), into partial fractions: $$\frac{1}{u(u+1)} = \frac{A}{u} + \frac{B}{u+1}$$ Multiplying both sides by \(u(u+1)\) to get rid of the denominators: $$1 = A(u+1) + Bu$$ Equating coefficients, we get: $$A + B = 0$$ $$A = 1$$ Solving for \(A\) and \(B\), we find that \(A = 1\) and \(B = -1\). Thus the integral becomes: $$\int \frac{du}{u(u+1)} = \int \left(\frac{1}{u} - \frac{1}{u+1}\right)du$$

Step 5: Integration and substitution back to \(x\)

We can now integrate the simplified expression: $$\int \left(\frac{1}{u} - \frac{1}{u+1}\right)du = \ln |u| - \ln |u + 1| + C$$ Since \(u = e^x\), we can write the result in terms of \(x\): $$\ln |e^x| - \ln |e^x + 1| + C = x - \ln (e^x + 1) + C$$ Therefore, the final result is: $$\int \frac{d x}{e^{x}+e^{2 x}} = x - \ln (e^x + 1) + C$$

Key Concepts

Change of Variables

When tackling integrals, especially ones with complex denominators, simplifying the expression can be a key step. One effective technique used is the **Change of Variables**. This method involves replacing a variable in the integral with a new one that's easier to work with. In our exercise, we started with an integral that included exponential functions: \[ \int \frac{d x}{e^{x}+e^{2 x}} \]To make this easier to handle, we introduce the substitution: let \( u = e^x \). This change of variables simplifies the expression dramatically because it transforms the denominator into a simpler polynomial, \( u^2 + u \).
After changing the variable, it's crucial to also change the differential, \(dx\), to \(du\). This requires finding the derivative of the substitution with respect to \(x\). Here, \( \frac{du}{dx} = e^x \), giving us the replacement \( dx = \frac{du}{e^x} \), which further helps in expressing the integral entirely in terms of \(u\). This technique reduces the complexity and often reveals further steps that better suit simpler forms or standard methods in calculus.

Partial Fraction Decomposition

Once a substitution simplifies an integral, you may still face expressions that require another technique to integrate: **Partial Fraction Decomposition**. This technique breaks down a complex rational expression into simpler fractions that are easier to integrate. Consider our new integrand: \[ \frac{1}{u(u+1)} \]The idea is to express this fraction as a sum of simpler fractions: \[ \frac{1}{u(u+1)} = \frac{A}{u} + \frac{B}{u+1} \]Here, \(A\) and \(B\) are constants determined by equating coefficients.
After multiplying through by the common denominator \(u(u+1)\), you get \[ 1 = A(u+1) + Bu \]. This simplifies to an equation system: - \( A + B = 0 \)- \( A = 1 \)Solving these, we find \( A = 1 \) and \( B = -1 \). This algebraic step allows us to rewrite the original integral into simpler, manageable parts: \[ \int \left(\frac{1}{u} - \frac{1}{u+1}\right) du \]. This decomposition is crucial because it transforms the integrand into two separate terms, each of which can be integrated directly using basic logarithmic identities.

Exponential Functions

Exponential functions, represented by expressions like \( e^x \), appear frequently in calculus. They have unique properties that make them a staple in mathematical analysis. Understanding their behavior is vital, especially when integrating functions that include exponentials. The integral in our exercise, \[ \int \frac{d x}{e^{x}+e^{2 x}} \], relies heavily on manipulating exponential forms. One of their key characteristics is that the derivative and the integral of \( e^x \) are both \( e^x \), a feature that simplifies many calculus problems. Additionally, exponentials often allow for substitutions, reducing complex expressions into more solvable terms, as seen with \( u = e^x \). This property enables us to change the integration variable, effectively transforming exponential terms into polynomials or simpler expressions that facilitate further techniques like partial fraction decomposition.

Indefinite Integral

The objective of many calculus exercises, including ours, is to find the **Indefinite Integral** of a function. This refers to integrating without defined limits, resulting in a general form that includes a constant of integration denoted by \(C\). The integral found in our problem was ultimately expressed as: \[ \int \frac{d x}{e^{x}+e^{2 x}} = x - \ln (e^x + 1) + C \].
In indefinite integrals, the result reflects a family of functions, all differing by a constant. This constant \(C\) accounts for all vertical shifts of the antiderivative on a graph since differentiation of a constant is zero. Indefinite integrals don't provide specific values but rather general solutions that describe any potential output derivative of the function given. Mastery over indefinite integration is essential for solving various problems in calculus, physics, and engineering, often laying the groundwork before applying specific bounds in definite integration.