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Chapter 15: Problem 24

\(\int \frac{\left(x^{2}-2\right) d x}{\left(x^{4}+5 x^{2}+4\right) \tan ^{-1}\left(\frac{x^{2}+2}{x}\right)}\) is (A) \(\log \left|\tan ^{-1} \sqrt{x+2}\right|+C\) (B) \(\log \left|\tan ^{-1}\left(x+\frac{2}{x}\right)\right|+C\) (C) \(\sin ^{-1}\left(\frac{x+2}{x}\right)+C\) (D) \(\tan ^{-1}\left(\frac{x+2}{x}\right)+C\)

Short Answer

Expert verified
The solution is option (D) \(\tan^{-1}\left(\frac{x+2}{x}\right) + C\).

Step by step solution

01

Analyze the Integral

The integral is \( \int \frac{(x^2 - 2) \, dx}{(x^4 + 5x^2 + 4) \tan^{-1}\left(\frac{x^2 + 2}{x}\right)} \). Here, the numerator function \(x^2 - 2\) appears to be related to the derivative of the argument of the arctangent in the denominator, \(\frac{x^2 + 2}{x}\).
02

Simplify the Expression

Note that the denominator \(x^4+5x^2+4\) factors as \((x^2+2)(x^2+2)\). Thus, both the numerator and the denominator have a common term \(x^2+2\). This indicates a potential simplification or substitution that utilizes properties of inverse trigonometric functions.
03

Substitute the Expression

Let \(u = \tan^{-1}\left(\frac{x^2 + 2}{x}\right)\). Then, \( \frac{du}{dx} = \frac{1}{1+\left(\frac{x^2 + 2}{x}\right)^2} \cdot \frac{d}{dx}\left(\frac{x^2 + 2}{x}\right) \). Simplifying gives \(du = \frac{x^2 - 2}{x^4 + 4x^2 + 4} \, dx\), which matches our integrand structure.
04

Solve the Integral Using Substitution

With the substitution \(u = \tan^{-1}\left(\frac{x^2 + 2}{x}\right)\), the integral becomes \(\int du = u + C\). Thus, the solution to the integral is \(\tan^{-1}\left(\frac{x^2 + 2}{x}\right) + C\).
05

Identify the Correct Answer

Comparing the obtained solution \(\tan^{-1}\left(\frac{x^2 + 2}{x}\right) + C\) with given options, it matches option (D).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculus
Integral calculus is a crucial part of calculus that focuses on the concept of integration. It is concerned with the process of finding the integral of functions, which, in simple terms, can be thought of as the reverse operation of differentiation. In this exercise, you are asked to find the integral of a given function, this means determining a function whose derivative matches the given expression, incorporating an arbitrary constant of integration \(C\).

Key points about integral calculus include:
  • Definite and Indefinite Integrals: An indefinite integral, like the one in this problem, provides a family of functions and is represented with the notation \(\int\). A definite integral, on the other hand, computes the area under a curve between two points.
  • Fundamental Theorem of Calculus: This theorem links differentiation and integration, two fundamental concepts in calculus, establishing that they are essentially inverse processes.
  • Application of Integrals: Integrals are used in various fields for calculating areas, volumes, central points, and many other essential calculations.
Understanding integral calculus helps solidify the connection between different calculus concepts and provides tools for solving a broad range of mathematical and real-world problems.
Inverse Trigonometric Functions
Inverse trigonometric functions are a class of functions that can undo the operations of the standard trigonometric functions like sine, cosine, and tangent. In this exercise, the solution involves the inverse tangent function, denoted as \(\tan^{-1}\). These functions are essential because they allow us to determine angles in right triangles when two sides are known.

Some important aspects of inverse trigonometric functions include:
  • Domain and Range: Inverse trig functions have specific domains and ranges, which often limit the angles they can evaluate to principal values. For example, for \(\tan^{-1}\), the range is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
  • Properties and Identities: These functions abide by several mathematical properties and identities that can be used to simplify integrals and solve equations.
  • Application in Calculus: In integration, inverse trig functions often appear as the antiderivatives of certain algebraic functions. In this exercise, \(\tan^{-1}\) is used to relate an intricate expression back to a familiar form, enabling easier integration.
Grasping how inverse trigonometric functions work enhances the understanding of angle measures and their relationships within triangles and circles.
Substitution Method
The substitution method, sometimes called \(u\)-substitution, is a powerful technique for finding integrals. It simplifies complex integrals into more manageable parts by introducing a new variable \(u\), which rephrases part of the integral. This is crucial in the given problem, as substitution clarifies the integration process.

Key aspects of the substitution method include:
  • Choosing the Right Substitution: A successful substitution usually involves identifying a function within the integrand whose derivative is also present in the integrand. In this exercise, substituting \(u = \tan^{-1}\left(\frac{x^2 + 2}{x}\right)\) effectively simplifies the integral.
  • Determining \(du\): Once a \(u\) is selected, you find \(du\), or the differential of \(u\), which replaces part of \(dx\) in the original integral.
  • Solving and Back-Substitution: After integration, substitute back the original variable to express the solution in terms of the initial variable, \(x\).
Using substitution method skillfully transforms complicated integrals into simpler ones, which makes solving them much more straightforward.

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Most popular questions from this chapter

If \(\frac{d y}{d x}=y+3 ; y>-3\) and \(\mathrm{y}(0)=2\), then y \((\ln 2)\) is equal to (A) 5 (B) 13 (C) 2 (D) 7

\(\int \frac{d x}{1-\cos ^{4} x}=-\frac{1}{2 \tan x}+\frac{k}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)+C\) where \(k=\) (A) \(\frac{1}{2}\) (B) \(-\frac{1}{2}\) (C) \(-1\) (D) 1

If \(I n=\int \frac{x^{n}}{\sqrt{x^{2}+a^{2}}} d x(n \geq 2)\), then In \(=\frac{x^{n-1} \sqrt{x^{2}+a^{2}}}{n}+k \operatorname{In}_{-2}\), where \(k=\) (A) \(\frac{a^{2}(1-n)}{n} I_{n-2}\) (B) \(\frac{a^{2}(n-1)}{n} I_{n-2}\) (C) \(\frac{a^{2}(n+1)}{n} I_{n-2}\) (D) none of these

\(\int \frac{x^{4}+1}{x^{6}+1} d x=\tan ^{-1} k_{1}-\frac{2}{3} \tan ^{-1} k_{2}+C\), where (A) \(k_{1}=x+\frac{1}{x}\) (B) \(k_{2}=x^{3}\) (C) \(k_{1}=x-\frac{1}{2}\) (D) \(k_{2}=x^{4}\)

\(\int \frac{1}{\left[(x-1)^{3}(x+2)^{5}\right]^{1 / 4}} d x\) is equal to (A) \(\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{14}+c\) (B) \(\frac{4}{3}\left(\frac{x+2}{x-1}\right)^{14}+c\) (C) \(\frac{1}{3}\left(\frac{x-1}{x+2}\right)^{14}+c\) (D) \(\frac{1}{3}\left(\frac{x+2}{x-1}\right)^{14}+c\)
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