Problem 70 - 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. $$\int \sqrt{e^{x}+1} d x(\text {Hint}: \text { Let } u=\sqrt{e^{x}+1} .)$$

Step-by-Step Solution

Answer

Question: Evaluate the integral $\int \sqrt{e^x + 1} dx$. Answer: $\frac{1}{2}(e^x-1) + C$

Step 1: Apply the change of variables

Let \(u = \sqrt{e^x + 1}\). Then we have \(u^2 = e^x + 1\). We need to differentiate with respect to \(x\) to obtain \(d u\) in terms of \(d x\). To do this, we first differentiate \(u^2\) with respect to \(u\), then multiply by \(d u / d x\): $$\frac{d}{d x}(u^2) = 2u \frac{d u}{d x} $$ Now, differentiate the right side of our equation \(u^2 = e^x + 1\) with respect to \(x\): $$\frac{d}{d x}(e^x + 1) = e^x$$ This gives us the equation: $$2u \frac{d u}{d x} = e^x$$ Now we can solve for \(d u\) in terms of \(d x\): $$d u = \frac{e^x}{2u} d x$$ Now substitute this new expression for \(du\) into the integral: $$\int \sqrt{e^x + 1} dx = \int u \frac{e^x}{2u} dx$$

Step 2: Simplify the integral

Now we can simplify the integral: $$\int u \frac{e^x}{2u} dx = \frac{1}{2} \int e^x dx$$

Step 3: Evaluate the simplified integral

Integrating \(\frac{1}{2}e^x\) with respect to \(x\) is straightforward: $$\frac{1}{2} \int e^x dx = \frac{1}{2}e^x + C$$ where \(C\) is the constant of integration.

Step 4: Substitute back the original variable

Now, we substitute back \(u = \sqrt{e^x + 1}\) to obtain the original variable: $$\frac{1}{2} e^x + C = \frac{1}{2} (u^2 - 1) + C$$ So, the final result is: $$\int \sqrt{e^x + 1} dx = \frac{1}{2} (\sqrt{e^x + 1}^2 - 1) + C = \frac{1}{2}(e^x-1) + C$$

Key Concepts

Change of Variables

The change of variables, also known as substitution, is a fundamental technique in integration. This method involves substituting a part of the integral with a new variable to simplify the integration process. By making a suitable substitution, we can transform a challenging integral into a simpler one. Here's how it works:

• Identify a part of the integral that can be substituted. In this context, this could be a function or expression that complicates the integration.
• Define a new variable that represents the selected part of the integral. For example, if we have an integral with a square root expression, setting it equal to a new variable can simplify the algebra involved.
• Differentiating both sides of the substitution equation helps to link the new variable back to the original variable.
• Replace the original expression in the integral with the new variable, transforming the integral into a more manageable form.

In the exercise, the expression under the square root, \(\sqrt{e^x + 1}\), was replaced by the variable \(u\), hence transforming the variable of integration. This transformation allowed the integral to be rewritten in terms of \(u\), simplifying the subsequent steps.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions. This method is particularly handy when integrating rational functions. Here's why and how you use it:

• The goal is to express a complex fraction as a sum of simpler fractions, which are easier to integrate individually.
• Start by factoring the denominator of the fraction, if possible. This helps in identifying the simpler fractions that the complex fraction will decompose into.
• Write the complex fraction as a sum of fractions with unknown coefficients. For linear factors in the denominator, use constants; for repeated or quadratic factors, use appropriate terms.
• Solve for the unknown coefficients by equating coefficients from the polynomial expansion or by substituting suitable values for the variable.

Although partial fraction decomposition wasn't explicitly utilized in the provided solution, appreciating its role is key in many integral evaluations, especially where the integrand is a complex rational function. In scenarios involving polynomial long division before partial fractions, these steps are critical.

Definite Integrals

Definite integrals represent the area under a curve within a specific interval. Unlike indefinite integrals, which produce a function plus an arbitrary constant \(C\), definite integrals yield a numerical value. Here's a breakdown of definite integrals:

• The limits of integration define the interval over which you integrate the function, often noted at the bottom and top of the integral symbol \(\int_{a}^{b}\).
• Definite integrals are evaluated using the Fundamental Theorem of Calculus, which links differentiation to integration.
• The process involves finding the antiderivative of the function and then evaluating it at the upper and lower limits, calculating their difference \(F(b) - F(a)\).
• This method does not result in "\(+C\)," the constant of integration; rather, it gives a specific numeric result for the area under \(f(x)\) from \(a\) to \(b\).

In the context of the solution provided, although the final integral wasn't evaluated as definite, knowing definite integrals is crucial. They appear frequently in applications dealing with real-world problems where quantifiable outcomes are required from the integration of a function.