Edexcel IAL Sample 2018 P4 (WMA14/01) Q3 Partial Fractions, Integration
\[f(x) = \frac{1}{x(3x-1)^2} = \frac{A}{x} + \frac{B}{(3x-1)} + \frac{C}{(3x-1)^2}\]
a. Find the values of the constants A, B and C
(4)
(4)
b. i. Hence, find \( \int f(x) \, dx \)
ii. Find \(\int_1^2 f(x) \, dx\), giving your answer in the form of \( a + \ln b \), where a and b are constants.
(6)
Solution to Part a: Expression in partial fractions
Key Concepts Used:
- Partial Fractions – Repeated Factors
Step-by-Step Working:
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Set \(\frac{1}{x(3x-1)^2}\) identical to \(\frac{A}{x} + \frac{B}{(3x-1)} + \frac{C}{(3x-1)^2}\)
\[\frac{1}{x(3x-1)^2} = \frac{A}{x} + \frac{B}{(3x-1)} + \frac{C}{(3x-1)^2}\]
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Let’s find the value of A, B, & C.
\[\frac{1}{x(3x-1)^2} = \frac{A(3x-1)^2 + Bx(3x-1) + Cx}{x(3x-1)^2}\]
\[1 = A(3x-1)^2 + Bx(3x-1) + Cx\]To find A, substitute \( x = 0 \):
\[1 = A(3 \times 0 – 1)^2 + B(0)(3 \times 0 – 1) + C(0)\]
\[1 = A(-1)^2\]
\[A = 1\]
\[1 = A(-1)^2\]
\[A = 1\]
To find C, substitute \( x = \frac{1}{3} \):
\[1 = A \left( 3 \times \frac{1}{3} – 1 \right)^2 + B \left( \frac{1}{3} \right) \left( 3 \times \frac{1}{3} – 1 \right) + C \left( \frac{1}{3} \right)\]
\[1 = A(0)^2 + B \left( \frac{1}{3} \right) (0) + C \left( \frac{1}{3} \right)\]
\[1 = C \left( \frac{1}{3} \right)\]
\[C = 3\]
\[1 = A(0)^2 + B \left( \frac{1}{3} \right) (0) + C \left( \frac{1}{3} \right)\]
\[1 = C \left( \frac{1}{3} \right)\]
\[C = 3\]
To find B, substitute \( x = 1 \):
\[1 = A(3 \times 1 – 1)^2 + B(1)(3 \times 1 – 1) + C(1)\]
\[1 = 4 + 2B + 3\]
\[2B = -6\]
\[B = -3\]
\[1 = 4 + 2B + 3\]
\[2B = -6\]
\[B = -3\]
\[\frac{1}{x(3x – 1)^2} = \frac{1}{x} – \frac{3}{(3x – 1)} + \frac{3}{(3x – 1)^2}\]
Final Answer:
\[\frac{1}{x(3x – 1)^2} = \frac{1}{x} – \frac{3}{(3x – 1)} + \frac{3}{(3x – 1)^2}\]
Solution to Part b(i): Integration
Key Concepts Used:
- Integrate Term-by-Term: Use standard integrals for each partial fraction
- Logarithmic Integration:
\[\int \frac{1}{ax + b} \, dx = \frac{1}{a} \ln|ax + b| + C\]
Step-by-Step Working:
1. Rewrite \( f(x) \):
\[f(x) = \frac{1}{x} – \frac{3}{3x – 1} + \frac{3}{(3x – 1)^2}\]
2. Integrate Each Term:
\[\int f(x) \, dx = \ln|x| – \ln|3x – 1| – \frac{1}{3x – 1} + C\]
Final Answer:
\[\ln|x| – \ln|3x – 1| – \frac{1}{3x – 1} + C\]
Solution to Part b(ii): Definite Integration
Key Concepts Used:
- Apply Limits: Substitute \( x = 1 \) and \( x = 2 \) into the antiderivative
- Simplify Logarithms: Use properties of logarithms to combine terms
Step-by-Step Working:
1. Apply the limits:
\[\left[\ln|x| – \ln|3x – 1| – \frac{1}{3x – 1}\right]_1^2\]
Compute at \( x = 2 \):
\[\ln2 – \ln5 – \frac{1}{5}\]
Compute at \( x = 1 \):
\[\ln1 – \ln2 – \frac{1}{2} = -\ln2 – \frac{1}{2}\]
Combine Results:
\[\left(\ln2 – \ln5 – \frac{1}{5}\right) – \left(-\ln2 – \frac{1}{2}\right) = 2\ln2 – \ln5 + \frac{3}{10}\]
\[\left(\ln2 – \ln5 – \frac{1}{5}\right) – \left(-\ln2 – \frac{1}{2}\right) = 2\ln2 – \ln5 + \frac{3}{10}\]
2. Simplify Logarithms:
\[2\ln2 – \ln5 = \ln4 – \ln5 = \ln\left(\frac{4}{5}\right)\]
Final Answer:
\[\frac{3}{10} + \ln\left(\frac{4}{5}\right)\]