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3. (a) Express \( \frac{5x + 3}{(2x - 3)(x + 2)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 6 - 2005 - Paper 6
Question 6
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 6 - 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 start with the form:
[ \frac{5x + 3}{(2x - 3)(x + 2)} = \frac{A}{2x - 3} + \frac{B}{x + 2} ]
Multiplying both sides by ( (2x - 3)(x + 2) ) gives:
[ 5x + 3 = A(x + 2) + B(2x - 3) ]
Expanding the right-hand side results in:
[ 5x + 3 = Ax + 2A + 2Bx - 3B ]
Combining like terms:
[ 5x + 3 = (A + 2B)x + (2A - 3B) ]
This gives us a system of equations:
- ( A + 2B = 5 )
- ( 2A - 3B = 3 )
Now we can solve for ( A ) and ( B ).
Using substitution or elimination, we find:
From equation 1: ( A = 5 - 2B ) and substituting into equation 2:
[ 2(5 - 2B) - 3B = 3 ] [ 10 - 4B - 3B = 3 ] [ 10 - 7B = 3 ] [ -7B = -7 \implies B = 1 ]
Substituting ( B = 1 ) back into equation 1:
[ A + 2(1) = 5 \implies A = 3 ]
Thus, we have: [ A = 3, \quad B = 1 ]
Therefore: [ \frac{5x + 3}{(2x - 3)(x + 2)} = \frac{3}{2x - 3} + \frac{1}{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
Now we can evaluate the integral:
[ \int \frac{5x + 3}{(2x - 3)(x + 2)} , dx = \int \left( \frac{3}{2x - 3} + \frac{1}{x + 2} \right) , dx ]
This can be split into two separate integrals:
[ \int \frac{3}{2x - 3} , dx + \int \frac{1}{x + 2} , dx ]
For the first integral, we have:
[ \int \frac{3}{2x - 3} , dx = \frac{3}{2} \ln |2x - 3| + C_1 ]
For the second integral:
[ \int \frac{1}{x + 2} , dx = \ln |x + 2| + C_2 ]
Combining these results, we get:
[ \int \frac{5x + 3}{(2x - 3)(x + 2)} , dx = \frac{3}{2} \ln |2x - 3| + \ln |x + 2| + C ]
Now, we need to find the definite integral from ( x = -1 ) to ( x = 3 ):
[ \left[ \frac{3}{2} \ln |2x - 3| + \ln |x + 2| \right]_{-1}^{3} ]
Evaluating this from ( x = -1 ) to ( x = 3 ):
At ( x = 3: ) [ \frac{3}{2} \ln |3| + \ln(5) = \frac{3}{2} \ln(3) + \ln(5) ]
At ( x = -1: ) [ \frac{3}{2} \ln(5) + \ln(1) = \frac{3}{2} \ln(5) ]
Thus, the result is: [ \left( \frac{3}{2} \ln(3) + \ln(5) \right) - \frac{3}{2} \ln(5) = \frac{3}{2} \ln(3) - \frac{1}{2} \ln(5) ]
Combining the logarithms gives: [ = \ln \left( \frac{3^{3/2}}{5^{1/2}} \right) = \ln \left( \frac{3\sqrt{3}}{\sqrt{5}} \right) = \ln \left( \frac{3 \sqrt{3}}{\sqrt{5}} \right) ]