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Express \( \frac{4x+3}{(x-1)^2} \) in the form \( \frac{A}{x-1} + \frac{B}{(x-1)^2} \) 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 Mechanics - Question 7 - 2019 - Paper 3
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![Express-\(-\frac{4x+3}{(x-1)^2}-\)-in-the-form-\(-\frac{A}{x-1}-+-\frac{B}{(x-1)^2}-\)--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 Mechanics-Question 7-2019-Paper 3.png](https://cdn.simplestudy.io/assets/backend/uploads/question/7198_question_7_3474.jpg)
Express \( \frac{4x+3}{(x-1)^2} \) in the form \( \frac{A}{x-1} + \frac{B}{(x-1)^2} \) Show that \[ \int_{3}^{4} \frac{4x+3}{(x-1)^2} dx = p + \ln q \] where \( p \... show full transcript
Worked Solution & Example Answer:Express \( \frac{4x+3}{(x-1)^2} \) in the form \( \frac{A}{x-1} + \frac{B}{(x-1)^2} \) 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 Mechanics - 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 desired form, we start by setting:
[ \frac{4x + 3}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}. ]
Multiplying through by ( (x-1)^2 ) gives:
[ 4x + 3 = A(x-1) + B. ]
Expanding the right-hand side:
[ 4x + 3 = Ax - A + B. ]
Grouping like terms results in:
[ 4x + 3 = Ax + (B - A). ]
Setting coefficients equal gives:
- For x: ( A = 4 )
- For constant: ( B - A = 3 ) ( \Rightarrow B = 3 + A, ) substituting ( A ) yields ( B = 3 + 4 = 7. )
Thus, we find ( A = 4 ) and ( B = 7. )
Step 2
Show that \( \int_{3}^{4} \frac{4x+3}{(x-1)^2} dx = p + \ln q \)
Answer
We begin by substituting our expression into the integral:
[ \int_{3}^{4} \frac{4x + 3}{(x-1)^2} dx = \int_{3}^{4} \left( \frac{4}{x-1} + \frac{7}{(x-1)^2} \right) dx. ]
This can be separated into two parts:
- ( \int \frac{4}{x-1} dx )
- ( \int \frac{7}{(x-1)^2} dx )
Calculating the first integral:
[ \int \frac{4}{x-1} dx = 4\ln|x-1| + C. ]
Evaluating from 3 to 4:
[ 4\ln|4-1| - 4\ln|3-1| = 4\ln(3) - 4\ln(2) = 4\ln\left( \frac{3}{2} \right). ]
Now for the second integral:
[ \int \frac{7}{(x-1)^2} dx = -\frac{7}{x-1} + C. ]
Evaluating from 3 to 4:
[ -\frac{7}{4-1} + \frac{7}{3-1} = -\frac{7}{3} + \frac{7}{2}. ]
Finding a common denominator gives:
[ -\frac{14}{6} + \frac{21}{6} = \frac{7}{6}. ]
Putting it all together, we have:
[ 4\ln\left( \frac{3}{2} \right) + \frac{7}{6}. ]
Thus, we can express the result in the form ( p + \ln q ) where ( p = \frac{7}{6} ) and ( q = \frac{3}{2}. )