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Magazine Chapter 6: Problem 7
Find the indefinite integral. $$ \int\left(x^{3}+x^{2}+x+1\right)\left(3 x^{4}+4 x^{3}+6 x^{2}+12 x+1\right)^{3} d x $$
Short Answer
Expert verified
\( \frac{1}{48}(3x^{4} + 4x^{3} + 6x^{2} + 12x + 1)^{4} + C \).
Step by step solution
01
Identify the Structure of the Integral
We need to find the indefinite integral \( \int (x^{3}+x^{2}+x+1)(3x^{4}+4x^{3}+6x^{2}+12x+1)^{3} \, dx \). Notice that the expression inside the integral has a polynomial \((3x^{4}+4x^{3}+6x^{2}+12x+1)\) raised to the power 3. This suggests the use of substitution, which might simplify the integration process.
02
Choose a Suitable Substitution
Let's choose \( u = 3x^{4} + 4x^{3} + 6x^{2} + 12x + 1 \), then the derivative will be \( du = (12x^{3} + 12x^{2} + 12x + 12) \, dx \) or \( du = 12(x^{3} + x^{2} + x + 1) \, dx \). This means \( dx = \frac{du}{12(x^{3} + x^{2} + x + 1)} \). Note how the factor of \( (x^{3} + x^{2} + x + 1) \) in the integrand matches this derivative, making this a strategic simplification.
03
Substitute and Simplify the Integral
Substitute \( u \) and \( dx \) into the integral:\[ \int (x^{3} + x^{2} + x + 1) (3x^{4} + 4x^{3} + 6x^{2} + 12x + 1)^{3} \, dx = \int (x^{3} + x^{2} + x + 1) u^{3} \left(\frac{du}{12(x^{3} + x^{2} + x + 1)}\right) \]This simplifies to\[ \frac{1}{12} \int u^{3} \, du \].
04
Integrate with Respect to u
Now, integrate \( u^{3} \) with respect to \( u \):\[ \int u^{3} \, du = \frac{u^{4}}{4} + C \]Therefore, the integral becomes\[ \frac{1}{12} \cdot \frac{u^{4}}{4} + C = \frac{1}{48} u^{4} + C \].
05
Substitute Back to x
Replace \( u \) with the original expression in terms of \( x \):\[ \frac{1}{48} (3x^{4} + 4x^{3} + 6x^{2} + 12x + 1)^{4} + C \].This gives us the indefinite integral in terms of \( x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a clever technique used to simplify the integration of complex functions. It's like tipping over the first domino to set off a chain reaction of simplification steps. Here's how it works:
- Identify the part of the integrand that could simplify the process. In our exercise, it's the polynomial \( (3x^4 + 4x^3 + 6x^2 + 12x + 1) \) raised to the third power.
- Select a substitution. Typically, this involves setting \( u \) equal to the part of the function that makes the derivative \( du \) manageable against the remaining terms. For instance, choosing \( u = 3x^4 + 4x^3 + 6x^2 + 12x + 1 \) works nicely since it correlates with \( x^3 + x^2 + x + 1 \).
- Differentiate your chosen substitution to find \( du \). This step determines exactly how \( du \) relates to \( dx \).
- Perform the substitution in the integral to transform everything into terms of \( u \), which makes the integral much more straightforward to evaluate.
Polynomial Integration
Polynomial integration involves finding the integral of polynomial expressions, which is a breeze once you master the basic rules. Let's break it down:
- If you have a polynomial like \( ax^n \), the integral is \( \frac{a}{n+1} x^{n+1} + C \), where \( C \) is the constant of integration. Simply increase the exponent by 1 and divide by this new exponent.
- Integration is performed on each term individually when the polynomial is composed of multiple terms. For example, each part of \( x^3 + x^2 + x + 1 \) would be integrated separately.
- The power rule is particularly handy with polynomials, allowing you to handle them term by term. This means even nested polynomials like the one raised to a power in our example can be split and handled conveniently.
- Keep track of constants throughout the integration process to avoid miscalculations. Remember: coefficients from differentiation and integration steps have a significant impact.
Integration Techniques
Understanding various integration techniques is crucial for tackling a variety of indefinite integrals. Integration isn't one-size-fits-all, and these methods help simplify complex expressions:
- Substitution method**: This technique is perfect for dealing with products of functions or complicated compositions, where substituting \( u \) simplifies the problem drastically, as shown in our exercise.
- Integration by parts**: Best used when the integrand is the product of two functions where substitution isn't feasible. It uses the formula \(\int u \, dv = uv - \int v \, du\).
- **Partial fraction decomposition**: Useful when dealing with rational functions, particularly for converting complex rational expressions into easier-to-integrate parts.
- **Trigonometric substitutions**: Handy if trigonometric identities can simplify the integrands, often transforming them into polynomial-like integrals.
