Solving Partial Fractions: A Comprehensive Guide

Partial fraction decomposition is an essential technique in algebra used to simplify complex rational expressions. This method is crucial in various mathematical calculations, including integration, and is widely used in engineering and physics. This article explores the step-by-step process of solving partial fractions and provides a detailed example for clarity.

Introduction to Partial Fractions

Partial fractions involve breaking down a complex rational function into simpler fractions. The main objective is to decompose a single rational expression into a sum of simpler rational expressions that are easier to handle.

Case Studies and Examples

The Problem: ( frac{x}{(x-1)^2(x 1)} )

Let's start by tackling the expression ( frac{x}{(x-1)^2(x 1)} ) . This expression can be decomposed using partial fractions, and the process involves finding constants A, B, and C such that the given expression can be written as a sum of simpler fractions.

Step 1: Set Up the Partial Fraction Decomposition

We express the given function as:

( frac{x}{(x-1)^2(x 1)} frac{A}{x-1} frac{B}{(x-1)^2} frac{C}{x 1} )

Step 2: Combine the Right-Hand Side into a Single Fraction

By combining the right-hand side into a single fraction, we obtain:

( frac{A(x-1)(x 1) B(x 1) C(x-1)^2}{(x-1)^2(x 1)} frac{x}{(x-1)^2(x 1)} )

Equating the numerators, we get:

( A(x^2 - 1) B(x 1) C(x^2 - 2x 1) x )

Step 3: Simplify and Solve for Constants

Expanding and simplifying the left-hand side, we have:

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

Rearranging terms, we get:

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

Now, we equate the coefficients of ( x^2 ), ( x ), and the constant term on both sides:

Coefficient of ( x^2 ): ( A C 0 ) …… (1) Coefficient of ( x ): ( B - 2C 1 ) …… (2) Constant term: ( B - A C 0 ) …… (3)

Step 4: Solving the System of Equations

From equation (1), we have ( A -C ).

Substituting ( A -C ) into equation (3), we get:

( B C C 0 ) implies ( B -2C ) …… (4)

Substituting ( B -2C ) into equation (2), we get:

( -2C - 2C 1 ) implies ( -4C 1 ) thus ( C -frac{1}{4} ).

From ( C -frac{1}{4} ), ( A frac{1}{4} ), and ( B -2 left( -frac{1}{4} right) frac{1}{2} ).

Step 5: Write the Final Partial Fraction Decomposition

The final partial fraction decomposition is:

( frac{x}{(x-1)^2(x 1)} frac{frac{1}{4}}{x-1} frac{frac{1}{2}}{(x-1)^2} - frac{frac{1}{4}}{x 1} )

Conclusion

Partial fractions are a powerful tool in algebra that simplify complex rational expressions. By systematically solving for the constants A, B, and C, we can decompose the given expression into simpler fractions. Understanding this process enhances problem-solving skills and facilitates more efficient calculations in various mathematical and scientific contexts.