Chapter 14: Problem 3
Find: \((a) \int \frac{3 d x}{x} \quad(x \neq 0)\) \((b) \int \frac{d x}{x-2} \quad(x \neq 2)\) (c) \(\int \frac{2 x}{x^{2}+3} d x\) (d) \(\int \frac{x}{3 x^{2}+5} d x\)
Short Answer
Expert verified
(a) \( 3\ln{|x|} + C \), (b) \( \ln{|x-2|} + C \), (c) \( \ln{|x^2+3|} + C \), (d) \( \frac{1}{6}\ln{|3x^2+5|} + C \).
Step by step solution
01
Understanding the Integral (a)
We have the integral \( \int \frac{3}{x} \, dx \). This is a standard form of integral for \( \int \frac{a}{x} \, dx = a \cdot \ln{|x|} + C \). Here, \( a = 3 \).
02
Solving Integral (a)
Apply the formula: \( \int \frac{3}{x} \, dx = 3 \cdot \ln{|x|} + C \). Thus, the solution for (a) is \( 3 \ln{|x|} + C \).
03
Understanding the Integral (b)
We have the integral \( \int \frac{1}{x-2} \, dx \). This follows the form \( \int \frac{1}{x-a} \, dx = \ln{|x-a|} + C \). Here, \( a = 2 \).
04
Solving Integral (b)
Apply the formula: \( \int \frac{1}{x-2} \, dx = \ln{|x-2|} + C \). Thus, the solution for (b) is \( \ln{|x-2|} + C \).
05
Understanding the Integral (c)
We have the integral \( \int \frac{2x}{x^2+3} \, dx \). Recognize that the derivative of \( x^2 + 3 \) is \( 2x \), so this fits \( \int \frac{f'(x)}{f(x)} \, dx = \ln{|f(x)|} + C \).
06
Solving Integral (c)
With \( f(x) = x^2 + 3 \) and \( f'(x) = 2x \), apply the formula: \( \int \frac{2x}{x^2+3} \, dx = \ln{|x^2+3|} + C \).
07
Understanding the Integral (d)
We have the integral \( \int \frac{x}{3x^2+5} \, dx \). Consider substitution \( u = 3x^2 + 5 \), so \( du = 6x \, dx \) or \( dx = \frac{du}{6x} \).
08
Substitution for Integral (d)
Substitute \( u = 3x^2 + 5 \) and adjust \( dx = \frac{du}{6x} \) back into the integral, leading to \( \int \frac{x}{u} \cdot \frac{du}{6x} \). This simplifies to \( \int \frac{1}{6} \cdot \frac{1}{u} \, du \).
09
Solving Integral (d)
Integrate \( \int \frac{1}{6} \cdot \frac{1}{u} \, du \), giving \( \frac{1}{6} \ln{|u|} + C \). Substituting back \( u = 3x^2 + 5 \), the integral \( \int \frac{x}{3x^2+5} \, dx = \frac{1}{6} \ln{|3x^2+5|} + C \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals are a fundamental concept in calculus that deal with the calculation of the area under a curve within specified limits. Unlike indefinite integrals, which include a constant of integration (C), definite integrals evaluate to a specific number. When you compute a definite integral, you're essentially finding the net area between the curve and the x-axis over a specific interval. For example, if you wanted to find the area under the curve of a function from point \(a\) to \(b\), you would use the notation \( \int_{a}^{b} f(x) \, dx \). This process involves using the Fundamental Theorem of Calculus, which relates differentiation and integration.
- The first part of the theorem states that if a function is continuous over [a, b], then the function has an antiderivative over that interval.
- The second part gives us a formula to evaluate the integral using the antiderivative: \[ F(b) - F(a) \] where \( F \) is any antiderivative of \( f \).
Indefinite Integrals
Indefinite integrals, on the other hand, are all about finding the general form of antiderivatives of a function. They do not evaluate to a specific number like definite integrals. Instead, they include a family of functions, represented with a \(+ C\) to indicate the constant of integration. For instance, when solving \( \int \frac{3}{x} \, dx \), we identify the antiderivative as \( 3 \ln{|x|} + C \). The process involves:
- Recognizing basic integral forms, such as \( \int \frac{1}{x} \, dx = \ln{|x|} + C \).
- Applying integration rules, like power rule or substitution, to simplify and solve the integral.
- Understanding each integral might need a specific technique, like recognizing a form or substituting variables.
Integration Techniques
Various integration techniques help simplify and solve different types of integrals that can't be solved by basic rules alone. Here, some useful techniques include:
- Substitution Method: This is used when an integral contains a function and its derivative. By substituting parts of the integral with a new variable, simplification becomes easier, such as in \( \int \frac{2x}{x^2+3} \, dx \), where substitution greatly simplifies the process.
- Partial Fractions: Ideal for dealing with rational functions. You decompose a complicated fraction into simpler parts that are easier to integrate.
- Integration by Parts: This technique is based on the product rule for differentiation. It's especially helpful with products of functions.