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Answer:

[tex]\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \boxed{ -\sqrt{4 - x^2} + C }[/tex]

General Formulas and Concepts:
Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]

Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]
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 Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Methods: U-Substitution and U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx[/tex]

Step 2: Integrate Pt. 1

Identify variables for u-substitution/u-solve.

  1. Set u:
    [tex]\displaystyle u = 4 - x^2[/tex]
  2. [u] Differentiate [Derivative Rules and Properties]:
    [tex]\displaystyle du = -2x \ dx[/tex]
  3. [du] Rewrite [U-Solve]:
    [tex]\displaystyle dx = \frac{-1}{2x} \ du[/tex]

Step 3: Integrate Pt. 2

  1. [Integral] Apply U-Solve:
    [tex]\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \int {\frac{-x}{2x\sqrt{u}}} \, du[/tex]
  2. [Integrand] Simplify:
    [tex]\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \int {\frac{-1}{2\sqrt{u}}} \, du[/tex]
  3. [Integral] Rewrite [Integration Property - Multiplied Constant]:
    [tex]\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \frac{-1}{2} \int {\frac{1}{\sqrt{u}}} \, du[/tex]
  4. [Integral] Apply Integration Rule [Reverse Power Rule]:
    [tex]\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = -\sqrt{u} + C[/tex]
  5. [u] Back-substitute:
    [tex]\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \boxed{ -\sqrt{4 - x^2} + C }[/tex]

∴ we have used u-solve (u-substitution) to find the indefinite integral.

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Learn more about Calculus: https://brainly.com/question/27746485

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

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