Question

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

(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

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, \( a = 3 \).

Solving Integral (a)

Apply the formula: \( \int \frac{3}{x} \, dx = 3 \cdot \ln{|x|} + C \). Thus, the solution for (a) is \( 3 \ln{|x|} + C \).

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, \( a = 2 \).

Solving Integral (b)

Apply the formula: \( \int \frac{1}{x-2} \, dx = \ln{|x-2|} + C \). Thus, the solution for (b) is \( \ln{|x-2|} + C \).

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 \( \int \frac{f'(x)}{f(x)} \, dx = \ln{|f(x)|} + C \).

Solving Integral (c)

With \( f(x) = x^2 + 3 \) and \( f'(x) = 2x \), apply the formula: \( \int \frac{2x}{x^2+3} \, dx = \ln{|x^2+3|} + C \).

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 \( dx = \frac{du}{6x} \).

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 \( \int \frac{1}{6} \cdot \frac{1}{u} \, du \).

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 \( \int \frac{x}{3x^2+5} \, dx = \frac{1}{6} \ln{|3x^2+5|} + C \).

Key concepts

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 \( f \).

Applying these principles allows us to calculate complex areas with precision.

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.

Mastering indefinite integrals is crucial for problems involving function growth and behavior.

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.

Choosing the right technique depends on the form of the integrand. Each method requires practice to identify which technique to apply for optimal simplification and integration of functions.