Problem 39 - Calculus: One and Several Variables

Question

Use \((8.4 .3)\) to calculate the integral.. $$\int \frac{1}{\left(x^{2}+1\right)^{3}} d x.$$

Step-by-Step Solution

Answer

The short answer is: \[ \int \frac{1}{\left(x^{2}+1\right)^{3}} d x = \frac{x}{4(x^2 + 1)^{2}} + \frac{3x}{8(x^2 + 1)} + \frac{3}{8} \arctan(x) + C \]

Step 1: Identify the formula to use

We will use the reduction formula: \[ \int \frac{1}{\left(x^{2}+1\right)^{n}} d x = \frac{x}{2(n - 1)(x^2 + 1)^{n - 1}} + \frac{2n - 3}{2(n - 1)}\int \frac{1}{(x^2 + 1)^{n - 1}} dx \]

Step 2: Apply the formula to the given integral

The given integral is: \[ \int \frac{1}{\left(x^{2}+1\right)^{3}} d x \] So we have \(n=3\). Apply the formula: \[ \int \frac{1}{\left(x^{2}+1\right)^{3}} d x = \frac{x}{2(3 - 1)(x^2 + 1)^{3 - 1}} + \frac{2(3) - 3}{2(3 - 1)}\int \frac{1}{(x^2 + 1)^{3 - 1}} dx \]

Step 3: Simplify the result

Simplify the expression: \[ \int \frac{1}{\left(x^{2}+1\right)^{3}} d x = \frac{x}{4(x^2 + 1)^{2}} + \frac{3}{4}\int \frac{1}{(x^2 + 1)^{2}} dx \] Now, we have the result in the form of the first term and a new integral.

Step 4: Solve the remaining integral

The remaining integral can be computed by using the same formula with \(n=2\): \[ \int \frac{1}{(x^2 + 1)^{2}} dx = \frac{x}{2(x^2 + 1)} + \frac{1}{2} \int \frac{1}{x^2 + 1} dx \] Now, the remaining integral is a well-known arctangent function: \[ \int \frac{1}{x^2 + 1} dx = \arctan(x) + C \]

Step 5: Combine the results

Now, substitute the result of the remaining integral back to the overall integral: \[ \int \frac{1}{\left(x^{2}+1\right)^{3}} d x = \frac{x}{4(x^2 + 1)^{2}} + \frac{3}{4} \left(\frac{x}{2(x^2 + 1)} + \frac{1}{2} \arctan(x) + C\right) \] Simplify the expression: \[ \int \frac{1}{\left(x^{2}+1\right)^{3}} d x = \frac{x}{4(x^2 + 1)^{2}} + \frac{3x}{8(x^2 + 1)} + \frac{3}{8} \arctan(x) + C \] This is the final result: \[ \int \frac{1}{\left(x^{2}+1\right)^{3}} d x = \frac{x}{4(x^2 + 1)^{2}} + \frac{3x}{8(x^2 + 1)} + \frac{3}{8} \arctan(x) + C \]

Key Concepts

Reduction Formula

The reduction formula is a powerful tool in calculus for simplifying complex integrals, particularly those involving powers of functions in the denominator. This formula reduces the power of the expression, allowing for iterative simplification. It takes advantage of patterns that occur when differentiating and integrating powers of binomials and trigonometric functions.
For example, consider the integral \(\int \frac{1}{(x^2+1)^n} \, dx\). The reduction formula for this is:

• \(\int \frac{1}{(x^2+1)^n} \, dx = \frac{x}{2(n-1)(x^2+1)^{n-1}} + \frac{2n-3}{2(n-1)}\int \frac{1}{(x^2+1)^{n-1}} \, dx\)

Here, \(n\) is the exponent of the term \((x^2 + 1)\). By appropriately substituting \(n\), you can break down a complex integral into more manageable parts, often leading to a basic, well-known integral which can be easily solved.

Definite and Indefinite Integrals

When studying calculus integration, it is crucial to differentiate between definite and indefinite integrals. The indefinite integral, also known as antiderivative, is the reverse process of differentiation. It is a function that gives all possible antiderivatives of a function, represented as \(\int f(x) \, dx = F(x) + C\), where \(C\) is the constant of integration.
An indefinite integral lacks bounds and represents a family of functions. In contrast, a definite integral calculates the area under the curve of \(f(x)\) from point \(a\) to point \(b\), denoted as \(\int_a^b f(x) \, dx\). It provides a numeric result, contrasting with the function result of an indefinite integral.
Understanding the difference is essential for applying correct methods and understanding the results of integration. Remember:

• Indefinite integrals provide a general form and include a (+ C) constant.
• Definite integrals have limits and provide a specific numerical value.

Trigonometric Substitution

Trigonometric substitution is a technique in integration where algebraic expressions are transformed into trigonometric ones to simplify calculations. This method is particularly useful when dealing with square roots or quadratic expressions.
The main idea is to replace variables with trigonometric functions, taking advantage of their properties, such as pythagorean identities \(\sin^2(x) + \cos^2(x) = 1\). Common substitutions include:

• \(x = a \sin(\theta)\)
• \(x = a \tan(\theta)\)
• \(x = a \sec(\theta)\)

By substituting \(x\) with a trigonometric function, you can transform the integral into an equivalent form that is often easier to solve, especially when the expression involves \((x^2 + a^2)\) or \((x^2 - a^2)\).
This method is integral (pun intended) to solving complex integrals in calculus.

Arctan Function

The arctan function, or the inverse tangent function, is crucial in integration, particularly with expressions involving \( \frac{1}{x^2 + 1} \). Its antiderivative is given by \(\int \frac{1}{x^2 + 1} \, dx = \arctan(x) + C\).
Understanding the arctan function is essential in solving integrals related to trigonometric and inverse trigonometric functions. This function helps express results in a form that is often more meaningful and standardized in calculus problems.
Here's what makes the arctan function notable:

• It provides the angle whose tangent is a given number.
• It is especially useful in integrals that result in expressions that fall into the tangent family.
• The range of the arctan function is typically \((-\pi/2, \pi/2)\), aligning with common angle values in trigonometry.

By understanding the behavior and properties of the arctan function, you can effectively solve and simplify complex integrals that appear frequently in calculus.