Answer:
B. [tex]\displaystyle \int {(9x^\Big{\frac{7}{2}} - 2x + 8x^\Big{-\frac{1}{2}})} \, dx[/tex]
General Formulas and Concepts:
Algebra I
Distributive Property
Exponents
- Exponential Property [Multiplying]: [tex]\displaystyle b^m \cdot b^n = b^{m + n}[/tex]
- Exponential Property [Rewrite]: [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]
- Exponential Property [Root Rewrite]: [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]
Calculus
Integration
U-Substitution
Step-by-step explanation:
Step 1: Define
Identify.
[tex]\displaystyle \int {\sqrt{x} \Big( 9x^3 - 2\sqrt{x} + \frac{8}{x} \Big)} \, dx[/tex]
Step 2: Find
- [Integrand] Rewrite [Exponential Property - Root Rewrite]: [tex]\displaystyle \int {\sqrt{x} \Big( 9x^3 - 2\sqrt{x} + \frac{8}{x} \Big)} \, dx = \int {x^\Big{\frac{1}{2}} \Big( 9x^3 - 2x^\Big{\frac{1}{2}} + \frac{8}{x} \Big)} \, dx[/tex]
- [Integrand] Rewrite [Exponential Property - Rewrite]: [tex]\displaystyle \int {\sqrt{x} \Big( 9x^3 - 2\sqrt{x} + \frac{8}{x} \Big)} \, dx = \int {x^\Big{\frac{1}{2}} \Big( 9x^3 - 2x^\Big{\frac{1}{2}} + 8x^{-1} \Big)} \, dx[/tex]
- [Integrand] Expand [Distributive Property]: [tex]\displaystyle \int {\sqrt{x} \Big( 9x^3 - 2\sqrt{x} + \frac{8}{x} \Big)} \, dx = \int {\Big( 9x^3x^\Big{\frac{1}{2}} - 2x^\Big{\frac{1}{2}}x^\Big{\frac{1}{2}} + 8x^{-1}x^\Big{\frac{1}{2}} \Big)} \, dx[/tex]
- [Integrand] Simplify [Exponential Rule - Multiplying]: [tex]\displaystyle \int {\sqrt{x} \Big( 9x^3 - 2\sqrt{x} + \frac{8}{x} \Big)} \, dx = \int {(9x^\Big{\frac{7}{2}} - 2x + 8x^\Big{-\frac{1}{2}})} \, dx[/tex]
∴ our answer is B.
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Learn more about integrals: https://brainly.com/question/20156869
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Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
