Problem 417 - Calculus Volume 1

Question

For the following exercises, find the antiderivatives for the functions. $$\int \frac{e^{x}}{\sqrt{e^{2 x}-1}}$$

Step-by-Step Solution

Answer

The antiderivative is \( x + \ln|1 + \sqrt{1 - e^{-2x}}| + C \).

Step 1: Recognize the Substitution

First, we recognize that the function within the integral is suitable for a substitution that simplifies the integral. Notice that the expression under the square root is \(e^{2x} - 1\). We can let \( u = e^x \).

Step 2: Compute the Derivatives

Differentiate the substitution \( u = e^x \) to find the expression for \( dx \) in terms of \( du \). Since \( \frac{du}{dx} = e^x \), we have \( du = e^x \, dx \). Hence, \( dx = \frac{du}{e^x} = \frac{du}{u} \).

Step 3: Substitute into the Integral

Substitute \( u = e^x \) and \( dx = \frac{du}{u} \) into the integral. The given integral becomes \( \int \frac{u}{\sqrt{u^2 - 1}} \cdot \frac{du}{u} \), which simplifies to \( \int \frac{du}{\sqrt{u^2 - 1}} \).

Step 4: Recognize the Result as a Standard Integral

The integral \( \int \frac{du}{\sqrt{u^2 - 1}} \) is a standard form that evaluates to \( \ln|u + \sqrt{u^2 - 1}| + C \), where \( C \) is the constant of integration.

Step 5: Substitute Back to Original Variable

Substitute back the original variable \( x \) for \( u \). Recall that \( u = e^x \), so the antiderivative in terms of \( x \) is \( \ln|e^x + \sqrt{e^{2x} - 1}| + C \). Simplify to \( x + \ln|1 + \sqrt{1 - e^{-2x}}| + C \).

Key Concepts

Substitution Method

The substitution method is a powerful tool in integral calculus used to simplify integrals. It's comparable to changing variables in equations to make them easier to solve. The idea is to transform a complex expression into a more manageable form.

In our example, the integral contains a function under a square root: \( e^{2x} - 1 \). Here's a clear way to handle this using substitution:

• Start by identifying parts of the integral that form a composite function.
• Choose a new variable (often \( u \)) for substitution. For our problem, letting \( u = e^x \) simplifies the square root expression.
• Derive the new differential: If \( u = e^x \), it follows that \( du = e^x \, dx \). Hence, \( dx = \frac{du}{u} \).
• Substitute \( u \) and \( dx \) into the integral. This transforms the integral into a simpler form.

Substitution bridges the gap between complex integrals and their resolutions, allowing easier computations and manipulation of functions.

Integral Calculus

Integral calculus is the branch of mathematics that deals with finding the antiderivative or integral of functions. It's the counterpart of differential calculus, which focuses on derivatives.

The main purpose of integral calculus is to determine quantities like area, volume, and accumulated change. When integrating, we're essentially summing up an infinite number of infinitely small quantities to get a total.

• Definite Integrals: Evaluate the integral over an interval, giving a number as a result.
• Indefinite Integrals: Find the general form of the antiderivative, including a constant of integration \( C \).
• Antiderivatives: Give a general form where the derivative would yield the original function.

In the problem, when we integrate \( \int \frac{du}{\sqrt{u^2 - 1}} \), we are finding the antiderivative. The integral calculus transforms a function into another function that describes the accumulation of its rate of change.

Standard Integrals

Standard integrals are well-known integral forms that involve common functions and are solved without detailed steps, as their solutions are memorized or derived in textbooks.

In the context of our exercise, after the substitution method, the integral becomes \( \int \frac{du}{\sqrt{u^2 - 1}} \). This is recognized as a standard form integral.

• Familiar Forms: Certain integrals recur frequently, such as \( \int \frac{dx}{x^2 + a^2} \), \( \int e^x \, dx \), and trigonometric integrals.
• Use of Tables: For many standard integrals, solutions might be found in tables or reference guides.
• Applying Knowledge: Recognizing standard integrals speeds up solving integrals once substitution or simplification is complete.

The integral \( \int \frac{du}{\sqrt{u^2 - 1}} \) evaluates to \( \ln|u + \sqrt{u^2 - 1}| + C \), a known result that matches the integral's standard form, allowing us to quickly reach the solution.