Problem 31 - Calculus: Early Transcendentals

Question

Evaluate the indefinite integral. \( \displaystyle \int \frac{(\arctan x)^2}{x^2 + 1} \, dx \)

Step-by-Step Solution

Answer

The indefinite integral is \( \frac{(\arctan x)^3}{3} + C \).

Step 1: Identify the Substitution

Notice that the denominator resembles the derivative of \( \arctan x \), which is \( \frac{1}{x^2 + 1} \). Thus, let's use substitution to simplify the integral. Let \( u = \arctan x \Rightarrow du = \frac{1}{x^2 + 1} \, dx \).

Step 2: Change Variables

Substitute the expressions for \( u \) and \( du \) into the integral. The integral becomes \( \int u^2 \, du \).

Step 3: Integrate with Respect to \( u \)

Since \( \int u^2 \, du \) is a simple power rule problem, integrate to find \( \frac{u^3}{3} + C \) where \( C \) is the constant of integration.

Step 4: Substitute Back to x

Replace \( u \) with \( \arctan x \) to return to the variable \( x \). This gives us the final integrated function: \( \frac{(\arctan x)^3}{3} + C \).

Key Concepts

Indefinite Integral

In calculus, understanding an indefinite integral is crucial as it represents the family of all antiderivatives of a function. When you see an integral symbol without specified limits, you are looking at an indefinite integral. Instead of finding a definite number as the result, you seek a general function. This function encapsulates all possible areas under the curve of the function you're integrating.

• An indefinite integral is often expressed as \( \int f(x) \, dx \).
• The result will include a constant of integration, denoted by \( C \).

This constant accounts for the infinite number of antiderivatives that differ by a constant. Imagine it like all the parallel lines on a graph, each being a different antiderivative, shifted vertically by a different constant value. This constant becomes crucial when performing further calculations, like solving initial value problems, or checking boundary conditions.

Substitution Method

The substitution method is a powerful and essential technique used in integration to simplify complicated integrals. When you encounter an integral that contains a composite function or resembles a known derivative, substitution can help streamline the process. Here's how it works:

• Identify a portion of the integrand that can be substituted with a simpler variable, typically \( u \).
• Express \( du \) in terms of \( dx \) to replace the differential.

In our example, recognizing that the denominator \( x^2 + 1 \) is the derivative of \( \arctan x \), we set \( u = \arctan x \) and find that \( du = \frac{1}{x^2 + 1} \, dx \). The substitution transforms the integral into an easier form, \( \int u^2 \, du \), allowing for straightforward application of integration rules. Once integrated, don't forget to substitute back to the original variable to complete the problem.

Power Rule

The power rule is a favored method for integrating terms that are polynomial in nature. It's simple yet applicable to a wide range of functions. Whenever you face an integral of a term like \( u^n \), where \( n \) is not equal to -1, the power rule provides the solution. The rule is stated as: \[ \int u^n \, du = \frac{u^{n+1}}{n+1} + C \] This rule is extremely efficient in handling polynomial functions. In our specific task, after applying the substitution method, the integral reduces to \( \int u^2 \, du \). Using the power rule, we find the antiderivative is \( \frac{u^3}{3} + C \). This highlights how integrals can often be reduced into forms where simple rules provide elegant solutions.

Trigonometric Integrals

Trigonometric integrals involve the integration of functions that include trigonometric expressions, which often require particular techniques depending on the complexity of the function. In the context of our problem, one part of the integrand indirectly relates to a trigonometric function through the inverse tangent, \( \arctan x \).

• Functions such as \( \sin x \), \( \cos x \), and their inverses often pop up in integration problems.
• Understanding the derivatives of these functions helps in spotting potential substitutions or manipulations.

In many instances, recognizing a trigonometric function or its derivative can guide the choice of substitution, as seen with \( u = \arctan x \) in our task. Mastery of these integrals can greatly expand your mathematical toolkit in solving a broad spectrum of calculus problems, particularly in applications involving angles and periodic phenomena.