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3. (a) Express \( \frac{5x + 3}{(2x - 3)(x + 2)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 5 - 2005 - Paper 6
Question 5
3. (a) Express \( \frac{5x + 3}{(2x - 3)(x + 2)} \) in partial fractions. (b) Hence find the exact value of \( \int \frac{5x + 3}{(2x - 3)(x + 2)} \, dx \), giving ... show full transcript
Worked Solution & Example Answer:3. (a) Express \( \frac{5x + 3}{(2x - 3)(x + 2)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 5 - 2005 - Paper 6
Step 1
Express \( \frac{5x + 3}{(2x - 3)(x + 2)} \) in partial fractions.
Answer
To express ( \frac{5x + 3}{(2x - 3)(x + 2)} ) in partial fractions, we assume:
[ \frac{5x + 3}{(2x - 3)(x + 2)} = \frac{A}{2x - 3} + \frac{B}{x + 2} ]
Multiplying through by the denominator ( (2x - 3)(x + 2) ) gives:
[ 5x + 3 = A(x + 2) + B(2x - 3) ]
Expanding the right-hand side:
[ 5x + 3 = Ax + 2A + 2Bx - 3B ]
Combining like terms:
[ 5x + 3 = (A + 2B)x + (2A - 3B) ]
From this, we can equate coefficients:
- ( A + 2B = 5 )
- ( 2A - 3B = 3 )
We can solve these equations simultaneously. Substituting ( x = -2 ) gives: [ 5(-2) + 3 = 0A + B, \text{ leading to } B = -7 ]
Substituting ( B = -7 ) into ( A + 2(-7) = 5 ):
[ A - 14 = 5 ]
[ A = 19 ]
Thus, we have: [ \frac{5x + 3}{(2x - 3)(x + 2)} = \frac{19}{2x - 3} - \frac{7}{x + 2} ]
Step 2
Hence find the exact value of \( \int \frac{5x + 3}{(2x - 3)(x + 2)} \, dx \), giving your answer as a single logarithm.
Answer
Using the partial fractions found previously:
[ \int \frac{5x + 3}{(2x - 3)(x + 2)} , dx = \int \left( \frac{19}{2x - 3} - \frac{7}{x + 2} \right) dx ]
This separates into two integrals:
[ = 19 \int \frac{1}{2x - 3} , dx - 7 \int \frac{1}{x + 2} , dx ]
Using the substitution ( u = 2x - 3 ) for the first integral:
[ \int \frac{1}{u} , \frac{du}{2} = \frac{1}{2} \ln |u| + C \Rightarrow = \frac{19}{2} \ln |2x - 3| ]
For the second integral, it integrates directly:
[ -7 \ln |x + 2| ]
Combining these gives:
[ \int \frac{5x + 3}{(2x - 3)(x + 2)} , dx = \frac{19}{2} \ln |2x - 3| - 7 \ln |x + 2| + C ]
To evaluate definite integrals over limits, substituting values if required. The final rearranged single logarithmic form can be:
[ = \ln \left| \frac{(2x - 3)^{19/2}}{(x + 2)^{7}} \right| + C ]