Evaluating Integrals: A Step-by-Step Guide with Algebraic Techniques

Integral evaluation is a fundamental part of calculus, and sometimes it requires a deep understanding of algebraic manipulation to simplify the integrand. This article will guide you through the process of evaluating a specific integral algebraically, using advanced techniques and substitutions.

Introduction to Integral Evaluation

Evaluating integrals often involves breaking down the integrand into simpler components through algebraic manipulations. This process can be both challenging and rewarding, as it often leads to elegant solutions. In this article, we will solve the integral:

The Integral in Focus

The integral in question is:

I (int_{1}^{sqrt{2}} frac{x^4}{x^2 - 1^2 - 1} ,mathrm{d}x)

Algebraic Rewriting of the Integral

To evaluate this integral, we start by rewriting the numerator and the denominator algebraically. The original integral can be rewritten as:

I (int_{1}^{sqrt{2}} frac{x^4}{x^4 - 2x^2 - 2} ,mathrm{d}x)

Further simplification leads to:

I (sqrt{2} - 1 int_{1}^{sqrt{2}} frac{2x^2 - 2}{x^4 - 2x^2 - 2} ,mathrm{d}x sqrt{2} - 1 int_{1}^{sqrt{2}} frac{2 - 2x^{-2}}{x^2 - 2 - 2x^{-2}} ,mathrm{d}x)

Completing the Square and Substitution

To proceed, we complete the square in the denominator:

x^2 - 2 - 2x^{-2} (x - x^{-1}sqrt{2})^2 - 2sqrt{2})

Considering the possible forms of completion, we identify:

x^2 - 2 - 2x^{-2} (x x^{-1}sqrt{2})^2 - 2sqrt{2})

From this, we need to decompose the fraction to allow for substitution. We find constants A and B such that:

2 - 2x^{-2} A(1 - x^{-2}sqrt{2}) B(1 x^{-2}sqrt{2}))

Solving for A and B, we get:

A (frac{sqrt{2} 1}{sqrt{2}}), B (frac{sqrt{2} - 1}{sqrt{2}})

Decomposition and Integration

The integral can now be decomposed and split into two parts:

[int_{1}^{sqrt{2}} frac{2 - 2x^{-2}}{x^2 - 2 - 2x^{-2}} ,mathrm{d}x frac{sqrt{2} - 1}{sqrt{2}} int_{1}^{sqrt{2}} frac{1 - x^{-2}sqrt{2}}{x^2 - 2 - 2x^{-2}} ,mathrm{d}x - frac{sqrt{2} 1}{sqrt{2}} int_{1}^{sqrt{2}} frac{1 x^{-2}sqrt{2}}{x^2 - 2 - 2x^{-2}} ,mathrm{d}x]

Each integral can be handled separately:

1. [int_{1}^{sqrt{2}} frac{1 - x^{-2}sqrt{2}}{x^2 - 2 - 2x^{-2}} ,mathrm{d}x int_{1}^{sqrt{2}} frac{dx - x^{-1}sqrt{2}}{x - x^{-1}sqrt{2}^2 - 2 - 2sqrt{2}} ,mathrm{d}x int_{-sqrt{2}-1}^{sqrt{2}-1} frac{1}{t^2 - 2sqrt{2} - 1} ,mathrm{d}t)

2. [int_{1}^{sqrt{2}} frac{1 x^{-2}sqrt{2}}{x^2 - 2 - 2x^{-2}} ,mathrm{d}x int_{1}^{sqrt{2}} frac{dx x^{-1}sqrt{2}}{x x^{-1}sqrt{2}^2 - 2 2sqrt{2}} ,mathrm{d}x int_{-sqrt{2}-1}^{sqrt{2}-1} frac{1}{t^2 2sqrt{2} - 1} ,mathrm{d}t]

Using symmetry, we simplify the second integral:

[int_{1}^{sqrt{2}} frac{1 x^{-2}sqrt{2}}{x^2 - 2 - 2x^{-2}} ,mathrm{d}x 2 int_{0}^{sqrt{2}-1} frac{1}{t^2 - 2sqrt{2} - 1} ,mathrm{d}t frac{1}{sqrt{2sqrt{2} - 1}} arctanleft(frac{t}{sqrt{2sqrt{2} - 1}}right) Bigg|_{0}^{sqrt{2} - 1}]

Final Solution

Combining the results, we have:

[I sqrt{2} - 1 left(0 - frac{1}{sqrt{2sqrt{2} - 1}} arctanleft(frac{t}{sqrt{2sqrt{2} - 1}}right) Bigg|_{0}^{sqrt{2} - 1}right) sqrt{sqrt{2} - 1} arctanleft(sqrt{frac{sqrt{2} - 1}{2}}right)]

Conclusion

This detailed step-by-step guide provides a comprehensive method for evaluating the given integral through algebraic manipulation, substitution, and symmetry. This technique is not only useful in solving integrals but also highlights the importance of algebraic skills in advanced calculus.