IB Mathematics AA Partial fractions Study Notes
IB Mathematics AA Partial fractions Study Notes
IB Mathematics AA Partial fractions Study Notes Offer a clear explanation of Partial fractions , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Partial fractions.
Partial Fractions
Partial Fractions
Definition:
Partial fractions involve expressing a rational function (a fraction where the numerator and denominator are both polynomials) as a sum of simpler rational expressions. This technique is especially useful for integration and simplifying algebraic expressions.
Types of Denominators and Decomposition Forms:
Distinct Linear Factors:
If the denominator can be factored into distinct linear terms, such as:
$ \frac{P(x)}{(x – a)(x – b)} = \frac{A}{x – a} + \frac{B}{x – b} $
Repeated Linear Factors:
For a repeated factor like \((x – a)^2\), decompose as:
$ \frac{P(x)}{(x – a)^2} = \frac{A}{x – a} + \frac{B}{(x – a)^2} $
Irreducible Quadratic Factors:
For a quadratic like \(x^2 + 1\) that can’t be factored, use:
$ \frac{P(x)}{x^2 + 1} = \frac{Ax + B}{x^2 + 1} $
Example
Express the following as partial fractions:
\( \frac{2x + 1}{x^2 + x – 2} \)
▶️ Answer/Explanation
Factor the denominator: \( x^2 + x – 2 = (x + 2)(x – 1) \)
So, decompose: $ \frac{2x + 1}{(x + 2)(x – 1)} = \frac{A}{x + 2} + \frac{B}{x – 1} $
Multiply through by the denominator : $ 2x + 1 = A(x – 1) + B(x + 2) $
$ 2x + 1 = Ax – A + Bx + 2B = (A + B)x + (-A + 2B) $
From \( x \): \( A + B = 2 \)
From constants: \( -A + 2B = 1 \)
Add both equations: \( A + B = 2 \)
\( -A + 2B = 1 \)
\( (A + B) + (-A + 2B) = 2 + 1 \Rightarrow 3B = 3 \Rightarrow B = 1 \)
\( A + B = 2 \): \( A + 1 = 2 \Rightarrow A = 1 \)
$ \frac{2x + 1}{x^2 + x – 2} = \frac{1}{x + 2} + \frac{1}{x – 1} $
Using Partial Fractions to Rearrange the Integrand
When integrating rational functions (fractions involving polynomials), it’s often easier to break them into simpler components using partial fractions. Once decomposed, each term can be integrated individually using basic integration rules.
Steps to Rearrange and Integrate Using Partial Fractions:
- Factor the denominator (if possible).
- Decompose</ the rational function into partial fractions based on the type of factors.
- Simplify the expression by solving for unknown constants.
- Integrate each partial fraction term separately using standard integral rules.
Example
Evaluate the integral:
\( \int \frac{2x + 1}{x^2 + x – 2} \, dx \)
▶️ Answer/Explanation
Factor the denominator: $ x^2 + x – 2 = (x + 2)(x – 1) $
Decompose the rational function: $ \frac{2x + 1}{(x + 2)(x – 1)} = \frac{A}{x + 2} + \frac{B}{x – 1} $
Multiply through by the denominator: $ 2x + 1 = A(x – 1) + B(x + 2) $
$ 2x + 1 = Ax – A + Bx + 2B = (A + B)x + (-A + 2B) $
Compare coefficients:
\( A + B = 2 \)
\( -A + 2B = 1 \)
From \( A + B = 2 \), \( A = 2 – B \)
\( – (2 – B) + 2B = 1 \Rightarrow -2 + B + 2B = 1 \Rightarrow 3B = 3 \Rightarrow B = 1 \)
\( A = 2 – 1 = 1 \)
$ \frac{2x + 1}{x^2 + x – 2} = \frac{1}{x + 2} + \frac{1}{x – 1} $
Integrate term-by-term: $ \int \frac{2x + 1}{x^2 + x – 2} \, dx = \int \frac{1}{x + 2} \, dx + \int \frac{1}{x – 1} \, dx $
$ = \ln|x + 2| + \ln|x – 1| + C $