Evaluating The Integral Of (t^4 + 1) / (t^6 + 1) From 1 To 2

Hey math enthusiasts! Today, we're diving deep into the fascinating world of calculus to tackle a definite integral that might seem a bit daunting at first glance. We're going to break down the steps to evaluate the integral ∫(from 1 to 2) (t^4 + 1) / (t^6 + 1) dt. So, grab your thinking caps and let's get started!

Understanding the Integral

Before we jump into the solution, let's take a moment to understand what we're dealing with. The integral ∫(from 1 to 2) (t^4 + 1) / (t^6 + 1) dt represents the area under the curve of the function f(t) = (t^4 + 1) / (t^6 + 1) between the limits t = 1 and t = 2. This type of problem often appears in advanced calculus courses and requires a solid understanding of integration techniques, particularly partial fraction decomposition and trigonometric substitutions.

Why This Integral Is Interesting

This integral isn't your run-of-the-mill problem. The integrand (t^4 + 1) / (t^6 + 1) doesn't have an immediately obvious antiderivative. This means we'll need to employ some clever algebraic manipulation and integration strategies to find the solution. The presence of higher powers of 't' in both the numerator and the denominator suggests that partial fraction decomposition might be a fruitful approach.

Breaking Down the Problem

To effectively solve this integral, we'll follow a structured approach:

1. Factor the denominator: We'll start by factoring the denominator (t^6 + 1) into manageable parts. This is a crucial step for partial fraction decomposition.
2. Partial fraction decomposition: We'll decompose the integrand into simpler fractions. This will make the integration process significantly easier.
3. Integrate each term: We'll integrate each of the resulting fractions. This might involve trigonometric substitutions or other integration techniques.
4. Evaluate the definite integral: Finally, we'll evaluate the antiderivative at the limits of integration (t = 1 and t = 2) and subtract to find the definite integral's value.

Step-by-Step Solution

Let's walk through each step in detail to conquer this integral.

1. Factoring the Denominator

The denominator is t^6 + 1. We can recognize this as a sum of cubes: (t²⁾3 + 1^3. Recall the sum of cubes factorization formula: a^3 + b^3 = (a + b)(a^2 - ab + b^2). Applying this, we get:

t^6 + 1 = (t^2 + 1)(t^4 - t^2 + 1)

However, the quartic term (t^4 - t^2 + 1) can be further factored. To do this, we can use a clever trick by adding and subtracting t^2:

t^4 - t^2 + 1 = t^4 + 2t^2 + 1 - 3t^2 = (t^2 + 1)^2 - (√3 t)^2

Now we have a difference of squares, which factors as:

(t^2 + 1)^2 - (√3 t)^2 = (t^2 + √3 t + 1)(t^2 - √3 t + 1)

So, the complete factorization of the denominator is:

t^6 + 1 = (t^2 + 1)(t^2 + √3 t + 1)(t^2 - √3 t + 1)

2. Partial Fraction Decomposition

Now that we've factored the denominator, we can express the integrand as a sum of partial fractions:

(t^4 + 1) / (t^6 + 1) = (At + B) / (t^2 + 1) + (Ct + D) / (t^2 + √3 t + 1) + (Et + F) / (t^2 - √3 t + 1)

This looks intimidating, but it's a crucial step. To find the constants A, B, C, D, E, and F, we multiply both sides by (t^6 + 1) and equate the coefficients of like powers of t. This will give us a system of linear equations to solve.

Multiplying both sides by (t^6 + 1) yields:

t^4 + 1 = (At + B)(t^2 + √3 t + 1)(t^2 - √3 t + 1) + (Ct + D)(t^2 + 1)(t^2 - √3 t + 1) + (Et + F)(t^2 + 1)(t^2 + √3 t + 1)

Expanding and collecting terms, we get a polynomial equation. By equating the coefficients of t^5, t^4, t^3, t^2, t, and the constant term, we obtain a system of six linear equations in six unknowns. Solving this system (which can be done using software or careful manual calculation) gives us the values:

A = 0, B = 1 C = 1/2, D = -√3/2 E = -1/2, F = -√3/2

Thus, our partial fraction decomposition is:

(t^4 + 1) / (t^6 + 1) = 1 / (t^2 + 1) + (1/2 t - √3/2) / (t^2 + √3 t + 1) + (-1/2 t - √3/2) / (t^2 - √3 t + 1)

3. Integrating Each Term

Now we integrate each term separately:

∫ [1 / (t^2 + 1)] dt = tan^(-1)(t)

For the other two terms, we'll need to complete the square in the denominators and use a substitution. Let's focus on the second term:

∫ [(1/2 t - √3/2) / (t^2 + √3 t + 1)] dt

Complete the square in the denominator:

t^2 + √3 t + 1 = (t + √3/2)^2 + 1/4

Now, we can rewrite the integral as:

∫ [(1/2 t - √3/2) / ((t + √3/2)^2 + 1/4)] dt

Using a substitution u = t + √3/2, du = dt, the integral becomes a bit more manageable. After further manipulation and integration techniques (which involve logarithmic and arctangent forms), we get:

(1/4) ln(t^2 + √3 t + 1) - (√3/2) tan^(-1)(2t + √3)

A similar process for the third term ∫ [(-1/2 t - √3/2) / (t^2 - √3 t + 1)] dt yields:

(-1/4) ln(t^2 - √3 t + 1) - (√3/2) tan^(-1)(2t - √3)

4. Evaluating the Definite Integral

Now we combine the results and evaluate the definite integral from 1 to 2:

∫(from 1 to 2) (t^4 + 1) / (t^6 + 1) dt = [tan^(-1)(t) + (1/4) ln(t^2 + √3 t + 1) - (√3/2) tan^(-1)(2t + √3) - (1/4) ln(t^2 - √3 t + 1) - (√3/2) tan^(-1)(2t - √3)] (from 1 to 2)

Plugging in the limits and simplifying (which involves some tedious algebra), we arrive at the final answer:

Answer: tan^(-1)(2) - (1/3)tan^(-1)(8) + π/3

Final Thoughts

Wow, that was quite the journey! We successfully evaluated a challenging definite integral by employing factoring, partial fraction decomposition, completing the square, and trigonometric substitutions. This problem highlights the importance of mastering various integration techniques and being persistent in problem-solving.

So, next time you encounter a seemingly complex integral, remember the strategies we discussed today. Break it down, stay organized, and don't be afraid to try different approaches. Happy integrating, guys!