Partial fractions
Algebra and functions
- A2
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Express $\dfrac{x + 7}{(x - 1)(x + 3)}$ in partial fractions.
3 marks
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Express $\dfrac{5x - 1}{(x - 1)(x + 2)}$ in partial fractions.
3 marks
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Express $\dfrac{7x - 1}{(x + 1)(2x - 1)}$ in partial fractions.
4 marks
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Express $\dfrac{3x + 1}{(x - 1)^2}$ in partial fractions.
4 marks
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Express $\dfrac{2x^2 + 5}{(x - 1)(x + 2)}$ in partial fractions.
5 marks
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Express $\dfrac{x^2 + 3x + 1}{x(x + 1)}$ in partial fractions.
5 marks
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Express $\dfrac{4x^2 - x + 3}{(x - 1)(x^2 + 2)}$ in partial fractions.
6 marks
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Given that $\dfrac{4x^2 - x + 3}{(x - 1)(x^2 + 2)} = \dfrac{2}{x - 1} + \dfrac{2x + 1}{x^2 + 2}$, find $\displaystyle \int \dfrac{4x^2 - x + 3}{(x - 1)(x^2 + 2)}\,dx$.
7 marks
Mark scheme
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Answer. $\dfrac{2}{x - 1} - \dfrac{1}{x + 3}$
- $\dfrac{A}{x - 1} + \dfrac{B}{x + 3}$.
- $x = 1$ gives $A = 2$, and $x = -3$ gives $B = -1$.
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Answer. $\dfrac{4}{3(x - 1)} + \dfrac{11}{3(x + 2)}$
- $x = 1$ gives $3A = 4$.
- $x = -2$ gives $-3B = -11$.
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Answer. $\dfrac{8}{3(x + 1)} + \dfrac{5}{3(2x - 1)}$
- $x = -1$ gives $-3A = -8$.
- $x = \dfrac{1}{2}$ gives $\dfrac{3}{2}B = \dfrac{5}{2}$.
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Answer. $\dfrac{3}{x - 1} + \dfrac{4}{(x - 1)^2}$
- Use $\dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^2}$.
- $A(x - 1) + B = 3x + 1$, so $B = 4$ and $A = 3$.
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Answer. $2 + \dfrac{7}{3(x - 1)} - \dfrac{13}{3(x + 2)}$
- The fraction is improper. Divide to get $2 + \dfrac{-2x + 9}{(x - 1)(x + 2)}$.
- Split $-2x + 9$ to get $\dfrac{7}{3(x - 1)} - \dfrac{13}{3(x + 2)}$.
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Answer. $1 + \dfrac{1}{x} + \dfrac{1}{x + 1}$
- Divide to get $1 + \dfrac{2x + 1}{x(x + 1)}$.
- $\dfrac{2x + 1}{x(x + 1)} = \dfrac{1}{x} + \dfrac{1}{x + 1}$.
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Answer. $\dfrac{2}{x - 1} + \dfrac{2x + 1}{x^2 + 2}$
- Use $\dfrac{A}{x - 1} + \dfrac{Bx + C}{x^2 + 2}$.
- $x = 1$ gives $A = 2$.
- Comparing coefficients gives $B = 2$ and $C = 1$.
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Answer. $2\ln|x - 1| + \ln(x^2 + 2) + \dfrac{1}{\sqrt{2}}\arctan\left(\dfrac{x}{\sqrt{2}}\right) + c$
- Split the second numerator: $2x + 1 = 2x + 1$.
- $\displaystyle \int \dfrac{2}{x - 1}\,dx = 2\ln|x - 1|$.
- $\displaystyle \int \dfrac{2x}{x^2 + 2}\,dx = \ln(x^2 + 2)$.
- $\displaystyle \int \dfrac{1}{x^2 + 2}\,dx = \dfrac{1}{\sqrt{2}}\arctan\left(\dfrac{x}{\sqrt{2}}\right)$.