Edexcel IAL - Pure Maths 4- 6.3 Partial Fractions Integration- Study notes - New syllabus
Edexcel IAL – Pure Maths 4- 6.3 Partial Fractions Integration -Study notes- New syllabus
Edexcel IAL – Pure Maths 4- 6.3 Partial Fractions Integration -Study notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 6.3 Partial Fractions Integration
Integration Using Partial Fractions
Partial fractions are used to simplify rational expressions so that they can be integrated easily. A rational expression is a fraction in which both the numerator and denominator are polynomials.
After decomposition into partial fractions, each term can be integrated using basic rules of integration.
Forms Required at IAL Level
| Type | Integral | Method |
| Linear denominator | \( \dfrac{2}{3x+5} \) | Substitution or simple form |
| Repeated linear factor | \( \dfrac{3}{(x-1)^2} \) | Rewrite using powers |
| Chain-rule type | \( \dfrac{x}{x^2+5} \) | Use substitution |
| Power of linear factor | \( \dfrac{2}{(2x-1)^4} \) | Use substitution |
Standard Integrals Needed
\( \displaystyle\int \dfrac{1}{ax+b}\,dx = \dfrac{1}{a}\ln|ax+b| + C \)
\( \displaystyle\int (x-a)^{-n}\,dx = \dfrac{(x-a)^{-n+1}}{-n+1} + C,\quad n \neq 1 \)
Example
Evaluate \( \displaystyle\int \dfrac{2}{3x+5}\,dx \).
▶️ Answer / Explanation
Let \( u = 3x + 5 \), so \( du = 3dx \).
\( \displaystyle\int \dfrac{2}{3x+5}\,dx = \dfrac{2}{3}\int \dfrac{1}{u}\,du \)
\( = \dfrac{2}{3}\ln|u| + C \)
\( = \dfrac{2}{3}\ln|3x+5| + C \)
Example
Evaluate \( \displaystyle\int \dfrac{3}{(x-1)^2}\,dx \).
▶️ Answer / Explanation
Rewrite the integrand:
\( \dfrac{3}{(x-1)^2} = 3(x-1)^{-2} \)
\( \int 3(x-1)^{-2}dx = 3 \int (x-1)^{-2}dx \)
\( = 3 \left( \dfrac{(x-1)^{-1}}{-1} \right) + C \)
\( = -\dfrac{3}{x-1} + C \)
Example
Evaluate \( \displaystyle\int \dfrac{x}{x^2+5}\,dx \).
▶️ Answer / Explanation
Let \( u = x^2 + 5 \), so \( du = 2x dx \).
\( \displaystyle\int \dfrac{x}{x^2+5}dx = \dfrac{1}{2}\int \dfrac{1}{u}du \)
\( = \dfrac{1}{2}\ln|u| + C \)
\( = \dfrac{1}{2}\ln(x^2+5) + C \)
Example
Evaluate \( \displaystyle\int \dfrac{2}{(2x-1)^4}\,dx \).
▶️ Answer / Explanation
Let \( u = 2x – 1 \), so \( du = 2\,dx \) and \( dx = \dfrac{1}{2}du \).
Substitute into the integral:
\( \displaystyle\int \dfrac{2}{(2x-1)^4}dx = \int \dfrac{2}{u^4}\cdot \dfrac{1}{2}du \)
\( = \int u^{-4}du \)
\( = \dfrac{u^{-3}}{-3} + C \)
\( = -\dfrac{1}{3u^3} + C \)
Substitute back:
\( -\dfrac{1}{3(2x-1)^3} + C \)