Math Problem Statement

integral( 4x^2 ((2/x^2 + (3/5x^4)) dx

Solution

To solve the integral:

4x2(2x2+35x4)dx\int 4x^2 \left( \frac{2}{x^2} + \frac{3}{5x^4} \right) \, dx

we'll proceed by simplifying the integrand first.

Step 1: Distribute 4x24x^2 to each term inside the parentheses

4x22x2+4x235x44x^2 \cdot \frac{2}{x^2} + 4x^2 \cdot \frac{3}{5x^4}

This gives: =8x2x2+12x25x4= \frac{8x^2}{x^2} + \frac{12x^2}{5x^4}

Step 2: Simplify each term

  1. For 8x2x2\frac{8x^2}{x^2}: 8x2x2=8\frac{8x^2}{x^2} = 8

  2. For 12x25x4\frac{12x^2}{5x^4}: 12x25x4=125x2\frac{12x^2}{5x^4} = \frac{12}{5x^2}

So the integral simplifies to: (8+125x2)dx\int \left( 8 + \frac{12}{5x^2} \right) \, dx

Step 3: Integrate term by term

  1. For 8dx\int 8 \, dx: 8dx=8x\int 8 \, dx = 8x

  2. For 125x2dx\int \frac{12}{5x^2} \, dx: Rewrite 125x2\frac{12}{5x^2} as 125x2\frac{12}{5} x^{-2}: 125x2dx=1251x=125x\int \frac{12}{5} x^{-2} \, dx = \frac{12}{5} \cdot \frac{-1}{x} = -\frac{12}{5x}

Step 4: Combine results

So, the integral is: 8x125x+C8x - \frac{12}{5x} + C

where CC is the constant of integration.


Would you like more details on any steps, or have any questions?


Here are some related questions you might find interesting:

  1. How do we approach integrals involving multiple terms in the denominator?
  2. What methods are used to simplify complex rational functions before integration?
  3. How does the power rule apply in integration, especially with negative exponents?
  4. How do we determine constants in indefinite integrals?
  5. What are common pitfalls when distributing terms in integration problems?

Tip: When integrating terms with xx in both the numerator and denominator, simplify each term independently before attempting to integrate.

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Math Problem Analysis

Mathematical Concepts

Integration
Algebraic Simplification
Indefinite Integrals

Formulas

Power Rule for Integration: ∫x^n dx = x^(n+1)/(n+1) + C, for n ≠ -1
Integral of x^(-n): ∫x^(-n) dx = -1/x for n = 2

Theorems

Power Rule for Integration
Linear Combination of Integrals

Suitable Grade Level

Grades 11-12