Partial Fraction Decomposition of Complex Rational Expressions: An In-Depth Tutorial

Understanding Complex Fractions and Their Decomposition

Partial fraction decomposition is a powerful technique used in algebra and calculus to break down complex rational expressions into simpler components. This method is particularly useful in simplifying integrations and solving differential equations. In this tutorial, we will guide you through the process of decomposing a specific complex rational expression, (frac{2x^2}{x-1x^{31}}).

Decomposing the Expression

The given expression is (frac{2x^2}{x-1x^{31}}). To begin, it's important to recognize that the denominator can be factored. Let's start by factoring the polynomial in the denominator:

(x-1x^{31} (x-1)(x 1)(x^2-x 1))

With this factorization, we can rewrite the original expression as:

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

Setting Up the Partial Fraction Decomposition

Next, we set up the partial fraction decomposition in a general form:

[frac{2x^2}{(x-1)(x 1)(x^2-x 1)} frac{A}{x-1} frac{B}{x 1} frac{Cx D}{x^2-x 1}]

We need to determine the constants (A), (B), (C), and (D).

Expanding and Combining Terms

To find (A), (B), (C), and (D), we first multiply both sides of the equation by the common denominator ((x-1)(x 1)(x^2-x 1)).

[2x^2 A(x 1)(x^2-x 1) B(x-1)(x^2-x 1) (Cx D)(x-1)(x 1)]

Expanding each term individually on the right side:

First term: (A(x 1)(x^2-x 1) A(x^3 x^2 - x x^2 x 1) Ax^3 2Ax^2 A) Second term: (B(x-1)(x^2-x 1) B(x^3 - x^2 x - x^2 x - 1) Bx^3 - 2Bx^2 2Bx - B) Third term: ((Cx D)(x-1)(x 1) (Cx D)(x^2-1) Cx^3 Dx^2 - Cx - D)

Combining all terms:

(2x^2 Ax^3 2Ax^2 A Bx^3 - 2Bx^2 2Bx - B Cx^3 Dx^2 - Cx - D)

Nested constants, collect like terms:

(2x^2 (A B C)x^3 (2A - 2B D)x^2 (2B - C)x (A - B - D))

Setting Up the System of Equations

Now, by comparing coefficients on both sides, we can set up a system of equations:

(A B C 0) (2A - 2B D 2) (2B - C 0) (A - B - D 0)

Solving the System of Equations

Let's solve these equations step by step:

From EQUATION 1: (A B C 0), we get (C -A - B).

Substitute (C) into EQUATION 3: (2B - C 0), then:

[2B - (-A - B) 0 Rightarrow 2B A B 0 Rightarrow A 3B 0 Rightarrow A -frac{3}{2}B]

Substitute (A) and (C) back into EQUATION 4: (A - B - D 0), then:

[-frac{3}{2}B - B - D 0 Rightarrow -frac{5}{2}B - D 0 Rightarrow D -frac{5}{2}B]

Substitute (A) and (D) into EQUATION 2: (2A - 2B D 2), then:

[2(-frac{3}{2}B) - 2B - frac{5}{2}B 2 Rightarrow -3B - 2B - frac{5}{2}B 2 Rightarrow -frac{11}{2}B 2 Rightarrow B -frac{2}{11} cdot 2 -frac{4}{11}]

Calculate (A) and (D) using (B -frac{4}{11}):

[A -frac{3}{2} times -frac{4}{11} frac{6}{11}]

[D -frac{5}{2} times -frac{4}{11} frac{10}{11}]

Calculate (C):

[C -A - B -frac{6}{11} - -frac{4}{11} -frac{2}{11}]

Final Partial Fraction Decomposition

Therefore, the partial fraction decomposition of (frac{2x^2}{(x-1)(x 1)(x^2-x 1)}) is:

[frac{2x^2}{(x-1)(x 1)(x^2-x 1)} frac{frac{6}{11}}{x-1} - frac{-frac{4}{11}}{x 1} - frac{-frac{2}{11}x frac{10}{11}}{x^2-x 1}]

For clarity, this can be written as:

[frac{2x^2}{(x-1)(x 1)(x^2-x 1)} frac{6}{11(x-1)} frac{4}{11(x 1)} - frac{-frac{2}{11}x frac{10}{11}}{x^2-x 1}]

This completes the partial fraction decomposition process. Understanding and mastering such techniques can significantly simplify complex algebraic expressions, making them easier to manage and work with in mathematics and engineering applications.