AI Partial Fractions Solver

Partial Fractions

Introduction to Partial Fractions

Partial fractions are a way to break down complex rational expressions into simpler ones, which can be easier to integrate or differentiate. This method is particularly useful in calculus and algebra when dealing with complex fractions.

Rational Expressions

A rational expression is a fraction where both the numerator and the denominator are polynomials. For example:
\[ \frac{2x + 3}{x^2 – x – 6} \]

Goal of Partial Fractions

The goal is to express a given rational expression as the sum of simpler fractions. For example:
\[ \frac{2x + 3}{x^2 – x – 6} = \frac{A}{x – 3} + \frac{B}{x + 2} \]

Steps to Decompose into Partial Fractions

1. Factor the Denominator:
– Factor the denominator of the rational expression.
– For example, factor \( x^2 – x – 6 \) as \( (x – 3)(x + 2) \).

2. Set Up the Partial Fractions:
– Write the rational expression as a sum of fractions with unknown coefficients.
– For \( \frac{2x + 3}{(x – 3)(x + 2)} \), we write:
\[ \frac{2x + 3}{(x – 3)(x + 2)} = \frac{A}{x – 3} + \frac{B}{x + 2} \]

3. Clear the Denominator:
– Multiply both sides by the factored denominator to get rid of the denominators.
– \( 2x + 3 = A(x + 2) + B(x – 3) \)

4. Solve for Coefficients:
– Expand and combine like terms.
– \( 2x + 3 = Ax + 2A + Bx – 3B \)
– Combine the \( x \)-terms and the constants:
\[ 2x + 3 = (A + B)x + (2A – 3B) \]
– Set up a system of equations by comparing coefficients:
– \( A + B = 2 \)
– \( 2A – 3B = 3 \)
– Solve the system of equations to find \( A \) and \( B \).

5. Write the Final Decomposition:
– Substitute the values of \( A \) and \( B \) back into the partial fractions.
– For example, if \( A = 1 \) and \( B = 1 \):
\[ \frac{2x + 3}{(x – 3)(x + 2)} = \frac{1}{x – 3} + \frac{1}{x + 2} \]

Types of Denominators

The approach to partial fractions varies slightly depending on the nature of the factors of the denominator:

1. Distinct Linear Factors:
– If the denominator has distinct linear factors, like \( (x – a)(x – b) \), use:
\[ \frac{P(x)}{(x – a)(x – b)} = \frac{A}{x – a} + \frac{B}{x – b} \]

2. Repeated Linear Factors:
– If the denominator has repeated linear factors, like \( (x – a)^2 \), use:
\[ \frac{P(x)}{(x – a)^2} = \frac{A}{x – a} + \frac{B}{(x – a)^2} \]

3. Irreducible Quadratic Factors:
– If the denominator has irreducible quadratic factors, like \( (x^2 + bx + c) \), use:
\[ \frac{P(x)}{x^2 + bx + c} = \frac{Ax + B}{x^2 + bx + c} \]

4. Combination of Factors:
– Combine the above methods if the denominator is a combination of different types of factors.

Example Problem

Decompose the following rational expression into partial fractions:
\[ \frac{3x^2 + 5x + 2}{(x – 1)(x + 2)(x + 3)} \]

Solution:
1. Factor the denominator: \( (x – 1)(x + 2)(x + 3) \).
2. Set up the partial fractions:
\[ \frac{3x^2 + 5x + 2}{(x – 1)(x + 2)(x + 3)} = \frac{A}{x – 1} + \frac{B}{x + 2} + \frac{C}{x + 3} \]
3. Clear the denominator:
\[ 3x^2 + 5x + 2 = A(x + 2)(x + 3) + B(x – 1)(x + 3) + C(x – 1)(x + 2) \]
4. Expand and combine like terms.
5. Solve the resulting system of equations to find \( A \), \( B \), and \( C \).
6. Substitute the values back to get the partial fraction decomposition.