Edexcel IAL January 2024 P4 (WMA14/01) Q2 Partial Fractions, Integration
Given that
\( \frac{3x+4}{(x-2)(2x+1)^2} \equiv \frac{A}{x-2} + \frac{B}{2x+1} + \frac{C}{(2x+1)^2} \)
a. find the values of the constants A, B and C.
b. Hence find the exact value of
\( \int_{7}^{12} \frac{3x+4}{(x-2)(2x+1)^2} \, dx \)
Giving your answer in the form of \( p \ln q + r \) where p, q, and r are rational numbers.
Solution to Part a: Find the values of A, B, & C.
Key Concepts Used:
- Partial Fractions – Repeated Factors
Step-by-Step Working:
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Set \( \frac{3x+4}{(x-2)(2x+1)^2} \) identical to \( \frac{A}{x-2} + \frac{B}{2x+1} + \frac{C}{(2x+1)^2} \)
\( \frac{3x+4}{(x-2)(2x+1)^2} = \frac{A}{x-2} + \frac{B}{2x+1} + \frac{C}{(2x+1)^2} \)\( 3x+4 = A(2x+1)^2 + B(x-2)(2x+1) + C(x-2) \)
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Let’s find the value of A, B, & C.
To find A, substitute x = 2.
\( 3(2) + 4 = A(2 \times 2 + 1)^2 + B(2-2)(2 \times 2 + 1) + C(2-2) \)
\( 10 = 25A \)
\( A = \frac{2}{5} \)
\( 10 = 25A \)
\( A = \frac{2}{5} \)
To find C, substitute \( x = -\frac{1}{2} \).
\( 3\left(-\frac{1}{2}\right) + 4 = A\left(2 \times -\frac{1}{2} + 1\right)^2 + B\left(-\frac{1}{2} – 2\right)\left(2 \times -\frac{1}{2} + 1\right) + C\left(-\frac{1}{2} – 2\right) \)
\( \frac{5}{2} = -\frac{5}{2}C \)
\( C = -1 \)
\( 3\left(-\frac{1}{2}\right) + 4 = A\left(2 \times -\frac{1}{2} + 1\right)^2 + B\left(-\frac{1}{2} – 2\right)\left(2 \times -\frac{1}{2} + 1\right) + C\left(-\frac{1}{2} – 2\right) \)
\( \frac{5}{2} = -\frac{5}{2}C \)
\( C = -1 \)
To find B, use equating coefficient method. For this, expand the brackets and collect the like terms.
\( 3x + 4 = A(2x + 1)^2 + B(x-2)(2x + 1) + C(x-2) \)
Substitute the values of A and C found.
\( 3x + 4 = \frac{2}{5}(2x + 1)^2 + B(x-2)(2x + 1) – 1(x-2) \)
\( 3x + 4 = \frac{2}{5}(4x^2 + 4x + 1) + B(2x^2 + x – 4x – 2) – x + 2 \)
\( 3x + 4 = \frac{8}{5}x^2 + \frac{8}{5}x + \frac{2}{5} + 2Bx^2 – 3Bx – 2B – x + 2 \)
\( 3x + 4 = \left(\frac{8}{5} + 2B\right)x^2 + \left(\frac{8}{5} – 3B – 1\right)x + \frac{2}{5} – 2B + 2 \)
Equating the constant terms on both sides of equation.
\( 4 = \frac{2}{5} – 2B + 2 \)
\( 3x + 4 = A(2x + 1)^2 + B(x-2)(2x + 1) + C(x-2) \)
Substitute the values of A and C found.
\( 3x + 4 = \frac{2}{5}(2x + 1)^2 + B(x-2)(2x + 1) – 1(x-2) \)
\( 3x + 4 = \frac{2}{5}(4x^2 + 4x + 1) + B(2x^2 + x – 4x – 2) – x + 2 \)
\( 3x + 4 = \frac{8}{5}x^2 + \frac{8}{5}x + \frac{2}{5} + 2Bx^2 – 3Bx – 2B – x + 2 \)
\( 3x + 4 = \left(\frac{8}{5} + 2B\right)x^2 + \left(\frac{8}{5} – 3B – 1\right)x + \frac{2}{5} – 2B + 2 \)
Equating the constant terms on both sides of equation.
\( 4 = \frac{2}{5} – 2B + 2 \)
\( \frac{8}{5} = -2B \)
\( B = -\frac{4}{5} \)
\( B = -\frac{4}{5} \)
3. Hence,
\( \frac{3x+4}{(x-2)(2x+1)^2} \equiv \frac{2}{5(x-2)} – \frac{4}{5(2x+1)} – \frac{1}{(2x+1)^2} \)
\( \frac{3x+4}{(x-2)(2x+1)^2} \equiv \frac{2}{5(x-2)} – \frac{4}{5(2x+1)} – \frac{1}{(2x+1)^2} \)
Final Answer:
\( \frac{3x+4}{(x-2)(2x+1)^2} \equiv \frac{2}{5(x-2)} – \frac{4}{5(2x+1)} – \frac{1}{(2x+1)^2} \)
Solution to Part b: Evaluating the Integral
Key Concepts Used:
- Integration Techniques
- \( \int \frac{1}{x-2} dx = \ln|x-2| \).
- \( \int \frac{1}{2x+1} dx = \frac{1}{2} \ln|2x+1| \).
- \( \int \frac{1}{(2x+1)^2} dx = -\frac{1}{2(2x+1)} \).
Step-by-Step Working:
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Integrate Each Term:
\( \frac{2}{5} \ln|x-2| – \frac{4}{5} \cdot \frac{1}{2} \ln|2x+1| – \frac{1}{(2x+1)^2} + C \)
Simplified:
\( \frac{2}{5} \ln|x-2| – \frac{2}{5} \ln|2x+1| – \frac{1}{2x+1} + C \)
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Evaluate Definite Integral from 7 to 12:
\( \left[ \frac{2}{5} \ln(x-2) – \frac{2}{5} \ln(2x+1) + \frac{1}{2x+1} \right]_{7}^{12} \)
- At \( x = 12 \):
\( \frac{2}{5} \ln 10 – \frac{2}{5} \ln 25 + \frac{1}{25} \)
- At \( x = 7 \):
\( \frac{2}{5} \ln 5 – \frac{2}{5} \ln 15 + \frac{1}{15} \)
- At \( x = 12 \):
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Combine Results:
\( \left( \frac{2}{5} \ln 10 – \frac{2}{5} \ln 25 + \frac{1}{25} \right) – \left( \frac{2}{5} \ln 5 – \frac{2}{5} \ln 15 + \frac{1}{15} \right) \)
Simplify logarithmic terms:
\( \frac{2}{5} (\ln 10 – \ln 25 – \ln 5 + \ln 15) + \left( \frac{1}{25} – \frac{1}{15} \right) \)Further simplification for using logarithm properties:
\( \frac{2}{5} (\ln 10 – \ln 25 + \ln 15 – \ln 5) + \left( \frac{1}{25} – \frac{1}{15} \right) \)
\( \frac{2}{5} \left( \ln \frac{10}{25} + \ln \frac{15}{5} \right) + \left( \frac{1}{25} – \frac{1}{15} \right) \)
\( \frac{2}{5} \left( \ln \frac{2}{5} + \ln 3 \right) + \left( \frac{1}{25} – \frac{1}{15} \right) \)
\( \frac{2}{5} \ln \left( \frac{2 \times 3}{5} \right) – \frac{1}{75} \)
\( \frac{2}{5} \ln \left( \frac{6}{5} \right) – \frac{1}{75} \)
Final Answer:
\( \frac{2}{5} \ln \left( \frac{6}{5} \right) – \frac{1}{75} \)