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Use a table of integrals to determine the following indefinite integrals. These integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table. $$\int \frac{d t}{t\left(t^{8}-256\right)}$$

Short Answer

Expert verified
Question: Evaluate the integral: $$\int \frac{d t}{t\left(t^8-256\right)}$$ Answer: The integral evaluates to: $$-\frac{1}{256}\ln{|t|}+C$$

Step by step solution

01

Simplify the given expression

We are given the following integral: $$\int \frac{d t}{t\left(t^{8}-256\right)}$$ We can rewrite the expression by factoring out the difference of two squares: $$\int \frac{d t}{t\left((t^4)^2- 16^2\right)}$$ Now, we can factor the difference of two squares $$\int \frac{d t}{t\left(t^4-16\right)\left(t^4+16\right)}$$
02

Apply Partial Fraction Decomposition

We can decompose the given rational function into simpler fractions: $$\frac{1}{t\left(t^4-16\right)\left(t^4+16\right)} =\frac{A}{t} + \frac{Bt^3 + Ct^2 + Dt + E}{t^4-16} + \frac{Ft^3 + Gt^2 + Ht + I}{t^4+16}$$ To determine the constants A, B, C, D, E, F, G, H, and I, first clear the fractions to obtain the following equation: $$1 = A \left(t^4-16\right)\left(t^4+16\right) + t(Bt^3 + Ct^2 + Dt + E)\left(t^4+16\right) + t(Ft^3 + Gt^2 + Ht + I)\left(t^4-16\right)$$
03

Determine Constants

Plug in suitable values of \(t\) to solve for the constants. Note that for a proper choice, it is easier to determine the constants. For example, 1. Plug in \(t = 0\) to solve for A: $$1 = A(-16)(16) \implies A = -\frac{1}{256}$$ Now, we can equate the coefficients of the powers of \(t\) in both sides of the equation. After comparing the coefficients, we find that B = 0, C = 0, D = 0, E = 0, F = 0, G = 0, H = 0, and I = 0. Hence, the partial fraction decomposition of the given rational function is: $$\frac{-1/256}{t}$$
04

Integrate

Integrate the partial fraction to determine the indefinite integral: $$\int \frac{-1/256}{t}dt=-\frac{1}{256}\int \frac{1}{t}dt$$ We know from the integral table that $$\int \frac{1}{t}dt = \ln{|t|}+C$$ So, the final answer becomes $$-\frac{1}{256}\ln{|t|}+C$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Fraction Decomposition
Partial Fraction Decomposition is an algebraic technique used particularly in calculus to break down complex rational expressions into simpler fractions, which can be more easily integrated. When integrating rational functions, especially those with polynomial expressions in the denominator, it's often beneficial to decompose them into parts with lower-degree polynomials.

Let's illustrate this with the exercise at hand. We have the integral of a rational function where the denominator is a product of \( t \) and two quartic polynomials. Our goal is to write this complex fraction as a sum of simpler fractions, each with its unique constant in the numerator and a factor of the original denominator in its own denominator. After setting up the partial fractions, we equalize them with the original expression and solve for the constants by equating coefficients or substituting strategic values for \( t \).

In practical scenarios, we look for constants when the denominator can be factored into linear or irreducible quadratic factors. However, in this case, due to the higher powers of \( t \), the constants in the other fractions turned out to be zero, leaving us with a single term for integration.
Completing the Square
Completing the Square is a method used to transform a quadratic equation into a perfect square trinomial, enabling us to solve the equation or evaluate an integral with greater ease. Essentially, it's about reorganizing and adding terms to an expression to create a square of a binomial. This technique can be particularly helpful when dealing with integrals involving quadratics that do not factor nicely.

In the context of our exercise, while completing the square was mentioned, it wasn't necessary since the given expression factored into a difference of squares nicely. However, if the expression in the denominator was slightly different and required this technique, we would've rewritten it in the form \( (t^2 + b)^2 - c^2 \) to facilitate the integration process.
Integration by Substitution
Integration by Substitution, also known as u-substitution, is a method analogous to the chain rule in differentiation. It is used when an integral contains a function and its derivative. By substituting a part of the integral (typically the inner function of a composition) with a new variable, we simplify the integration process.

In this particular problem, although u-substitution wasn't directly applied, it is worth noting that if the integral had a more complicated composition of functions, this technique would likely be our method of choice. For example, if the denominator had an expression like \( (t^2 + 1)^2 \) rather than simple polynomials, u-substitution would be the appropriate approach to simplify the integral before finding a solution.
Difference of Two Squares
The Difference of Two Squares is a fundamental algebraic pattern that states that two terms squared and subtracted can be factored into the product of the sum and difference of those two terms, as a product of two binomials. The formula is given by \( a^2 - b^2 = (a + b)(a - b) \).

In our exercise, this pattern is visible where the integrand's denominator features a term \( t^8 - 256 \) which can be viewed as \( (t^4)^2 - 16^2 \) and thus factored into \( (t^4 + 16)(t^4 - 16) \). This step is critical because it opens the path to the partial fraction decomposition that follows, allowing us to express the integrand in a form that is simpler to integrate. Understanding the difference of squares enables us to tackle a wide range of integrals involving quadratic terms or higher powers that exhibit this pattern.

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