Respuesta :

Answer:

Diverges

Step-by-step explanation:

Here an improper integral is given.The limits are 1 to infinity

The function to be integrated is

[tex]\frac{4x+6}{x^2+3x}[/tex]

Let us decompose this into partial fractions

[tex]\frac{4x+6}{x^2+3x}[/tex]=[tex]\frac{A}{x}+\frac{B}{x+3}[/tex]

Solving we get

A=B=2

Hence integral would be = 2 ln x + 2 ln (x+3)

When we substitute infinity here, this becomes very large thus the integral diverges.

Space

Answer:

The improper integral diverges.

[tex]\displaystyle \int\limits^{\infty}_1 {\frac{4x + 6}{x^2 + 3x}} \, dx = \infty[/tex]

General Formulas and Concepts:
Calculus

Limits

Limit Rule [Variable Direct Substitution]:
[tex]\displaystyle \lim_{x \to c} x = c[/tex]

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:
[tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Derivative Property [Addition/Subtraction]:
\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Method: U-Substitution

Improper Integral:
[tex]\displaystyle \int\limits^{\infty}_a {f(x)} \, dx = \lim_{b \to \infty} \int\limits^b_a {f(x)} \, dx[/tex]

Step-by-step explanation:

Step 1: Define

Identify.

[tex]\displaystyle \int\limits^{\infty}_1 {\frac{4x + 6}{x^2 + 3x}} \, dx[/tex]

Step 2: Integrate Pt. 1

  1. [Integral] Rewrite [Improper Integral]:
    [tex]\displaystyle \int\limits^{\infty}_1 {\frac{4x + 6}{x^2 + 3x}} \, dx = \lim_{b \to \infty} \int\limits^{b}_1 {\frac{4x + 6}{x^2 + 3x}} \, dx[/tex]

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

  1. Set u:
    [tex]\displaystyle u = x^2 + 3x[/tex]
  2. [u] Differentiate [Derivative Properties and Rules]:
    [tex]\displaystyle u = x^2 + 3x[/tex]
  3. [Bounds] Swap:
    [tex]\displaystyle \left \{ {{x = b \rightarrow u = b^2 + 3b} \atop {x = 1 \rightarrow u = 4}} \right.[/tex]

Step 4: Integrate Pt. 3

  1. [Integrand] Rewrite:
    [tex]\displaystyle \int\limits^{\infty}_1 {\frac{4x + 6}{x^2 + 3x}} \, dx = \lim_{b \to \infty} \int\limits^{b}_1 {\frac{2(2x + 3)}{x^2 + 3x}} \, dx[/tex]
  2. [Integral] Apply Integration Method [U-Substitution]:
    [tex]\displaystyle \int\limits^{\infty}_1 {\frac{4x + 6}{x^2 + 3x}} \, dx = \lim_{b \to \infty} \int\limits^{b^2 + 3b}_4 {\frac{2}{u}} \, du[/tex]
  3. [Integral] Rewrite [Integration Property - Multiplied Constant]:
    [tex]\displaystyle \int\limits^{\infty}_1 {\frac{4x + 6}{x^2 + 3x}} \, dx = \lim_{b \to \infty} 2 \int\limits^{b^2 + 3b}_4 {\frac{1}{u}} \, du[/tex]
  4. [Integral] Apply Logarithmic Integration:
    [tex]\displaystyle \int\limits^{\infty}_1 {\frac{4x + 6}{x^2 + 3x}} \, dx = \lim_{b \to \infty} 2 \ln |u| \bigg| \limits^{b^2 + 3b}_4[/tex]
  5. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
    [tex]\displaystyle \int\limits^{\infty}_1 {\frac{4x + 6}{x^2 + 3x}} \, dx = \lim_{b \to \infty} 2 \ln \bigg( \frac{|b^2 + 3b|}{4} \bigg)[/tex]
  6. [Limit] Evaluate [Limit Rule - Variable Direct Substitution]:
    [tex]\displaystyle \int\limits^{\infty}_1 {\frac{4x + 6}{x^2 + 3x}} \, dx = 2 \ln \bigg( \frac{|\infty^2 + 3(\infty)|}{4} \bigg)[/tex]
  7. Simplify:
    [tex]\displaystyle \int\limits^{\infty}_1 {\frac{4x + 6}{x^2 + 3x}} \, dx = \infty[/tex]

∴ the improper integral tends to ∞ and is divergent.

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Learn more about improper integrals: https://brainly.com/question/14413608

Learn more about calculus: https://brainly.com/question/23558817

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Topic: AP Calculus BC (Calculus I + II)

Unit: Integration

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