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7 (a) Express \( \frac{4x+3}{(x-1)^2} \) in the form \( \frac{A}{x-1} + \frac{B}{(x-1)^2} \) 7 (b) Show that \[ \int_{3}^{4} \frac{4x + 3}{(x-1)^{2}} \, dx = p + \ln q \] where \( p \) and \( q \) are rational numbers. - AQA - A-Level Maths Pure - Question 7 - 2019 - Paper 3
Question 7
![7-(a)-Express-\(-\frac{4x+3}{(x-1)^2}-\)-in-the-form-\(-\frac{A}{x-1}-+-\frac{B}{(x-1)^2}-\)---7-(b)-Show-that---\[-\int_{3}^{4}-\frac{4x-+-3}{(x-1)^{2}}-\,-dx-=-p-+-\ln-q-\]---where-\(-p-\)-and-\(-q-\)-are-rational-numbers.-AQA-A-Level Maths Pure-Question 7-2019-Paper 3.png](https://cdn.simplestudy.io/assets/backend/uploads/question/7216_question_7_8050.jpg)
7 (a) Express \( \frac{4x+3}{(x-1)^2} \) in the form \( \frac{A}{x-1} + \frac{B}{(x-1)^2} \) 7 (b) Show that \[ \int_{3}^{4} \frac{4x + 3}{(x-1)^{2}} \, dx = p +... show full transcript
Worked Solution & Example Answer:7 (a) Express \( \frac{4x+3}{(x-1)^2} \) in the form \( \frac{A}{x-1} + \frac{B}{(x-1)^2} \) 7 (b) Show that \[ \int_{3}^{4} \frac{4x + 3}{(x-1)^{2}} \, dx = p + \ln q \] where \( p \) and \( q \) are rational numbers. - AQA - A-Level Maths Pure - Question 7 - 2019 - Paper 3
Step 1
Express \( \frac{4x+3}{(x-1)^2} \) in the form \( \frac{A}{x-1} + \frac{B}{(x-1)^2} \)
Answer
To express ( \frac{4x+3}{(x-1)^2} ) in the required form, we start by setting:
[ \frac{4x+3}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2} ]
Multiplying both sides by ( (x-1)^2 ) gives:
[ 4x + 3 = A(x-1) + B ]
Next, we expand the right side:
[ 4x + 3 = Ax - A + B ]
Rearranging terms gives:
[ 4x + 3 = Ax + (B - A) ]
By comparing coefficients:
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For the coefficient of ( x ), we have ( A = 4 ).
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For the constant term, we have ( B - A = 3 ), so substituting ( A ):
[ B - 4 = 3, \quad B = 7 ]
Thus, we have: [ A = 4, \quad B = 7 ]
Step 2
Show that \( \int_{3}^{4} \frac{4x + 3}{(x-1)^{2}} \, dx = p + \ln q \)
Answer
Now we integrate the expression:
[ \int \frac{4x + 3}{(x-1)^{2}} , dx ]
Using the results from part (a), we can rewrite the integrand:
[ \int \left( \frac{4}{x-1} + \frac{7}{(x-1)^{2}} \right) , dx ]
This can be split into two separate integrals:
[ \int \frac{4}{x-1} , dx + \int \frac{7}{(x-1)^{2}} , dx ]
Integrating each term gives:
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For the first integral: [ 4 \ln |x - 1| ]
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For the second integral (using ( \int x^{-n} , dx = \frac{x^{-n + 1}}{-n + 1} ) for ( n = 2 )): [ -\frac{7}{x-1} ]
Thus, the total integral from 3 to 4 becomes:
[ \left[ 4 \ln |x - 1| - \frac{7}{x - 1} \right]_{3}^{4} ]
Evaluating these limits:
At ( x = 4 ): [ 4 \ln |4 - 1| - \frac{7}{4 - 1} = 4 \ln 3 - \frac{7}{3} ]
At ( x = 3 ): [ 4 \ln |3 - 1| - \frac{7}{3 - 1} = 4 \ln 2 - \frac{7}{2} ]
Combining these gives:
[ \left( 4 \ln 3 - \frac{7}{3} \right) - \left( 4 \ln 2 - \frac{7}{2} \right) ]
This simplifies to: [ 4 \ln 3 - 4 \ln 2 + \frac{7}{2} - \frac{7}{3} ]
The final answer is presented in the form ( p + \ln q ) where ( p ) and ( q ) are rational numbers and can be derived from appropriate simplification.