Partial fractions after substitution

A question is this type if and only if it requires first applying a substitution to transform the integral, then using partial fractions to complete the integration.

8 questions · Standard +0.5

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CAIE P3 2024 November Q11
14 marks Standard +0.8
11 Let \(\mathrm { f } ( x ) = \frac { 2 \mathrm { e } ^ { 2 x } } { \mathrm { e } ^ { 2 x } - 3 \mathrm { e } ^ { x } + 2 }\).
  1. Find \(\mathrm { f } ^ { \prime } ( x )\) and hence find the exact coordinates of the stationary point of the curve with equation \(y = \mathrm { f } ( x )\).
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  2. Use the substitution \(u = e ^ { x }\) and partial fractions to find the exact value of \(\int _ { \ln 3 } ^ { \ln 5 } \mathrm { f } ( x ) \mathrm { d } x\). Give your answer in the form \(\ln a\), where \(a\) is a rational number in its simplest form.
    If you use the following page to complete the answer to any question, the question number must be clearly shown.
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Edexcel C4 2013 June Q5
10 marks Standard +0.3
  1. (a) Use the substitution \(x = u ^ { 2 } , u > 0\), to show that
$$\int \frac { 1 } { x ( 2 \sqrt { x } - 1 ) } \mathrm { d } x = \int \frac { 2 } { u ( 2 u - 1 ) } \mathrm { d } u$$ (b) Hence show that $$\int _ { 1 } ^ { 9 } \frac { 1 } { x ( 2 \sqrt { x } - 1 ) } \mathrm { d } x = 2 \ln \left( \frac { a } { b } \right)$$ where \(a\) and \(b\) are integers to be determined.
Edexcel C4 2018 June Q3
14 marks Standard +0.3
3. (i) Given that $$\frac { 13 - 4 x } { ( 2 x + 1 ) ^ { 2 } ( x + 3 ) } \equiv \frac { A } { ( 2 x + 1 ) } + \frac { B } { ( 2 x + 1 ) ^ { 2 } } + \frac { C } { ( x + 3 ) }$$
  1. find the values of the constants \(A , B\) and \(C\).
  2. Hence find $$\int \frac { 13 - 4 x } { ( 2 x + 1 ) ^ { 2 } ( x + 3 ) } \mathrm { d } x , \quad x > - \frac { 1 } { 2 }$$ (ii) Find $$\int \left( \mathrm { e } ^ { x } + 1 \right) ^ { 3 } \mathrm {~d} x$$ (iii) Using the substitution \(u ^ { 3 } = x\), or otherwise, find $$\int \frac { 1 } { 4 x + 5 x ^ { \frac { 1 } { 3 } } } \mathrm {~d} x , \quad x > 0$$
Edexcel P4 2021 June Q4
8 marks Standard +0.8
4. Use algebraic integration and the substitution \(u = \sqrt { x }\) to find the exact value of $$\int _ { 1 } ^ { 4 } \frac { 10 } { 5 x + 2 x \sqrt { x } } \mathrm {~d} x$$ Write your answer in the form \(4 \ln \left( \frac { a } { b } \right)\), where \(a\) and \(b\) are integers to be found.
(Solutions relying entirely on calculator technology are not acceptable.)
OCR C4 Q9
14 marks Standard +0.3
9. (i) Show that the substitution \(u = \sin x\) transforms the integral $$\int \frac { 6 } { \cos x ( 2 - \sin x ) } d x$$ into the integral $$\int \frac { 6 } { \left( 1 - u ^ { 2 } \right) ( 2 - u ) } \mathrm { d } u$$ (ii) Express \(\frac { 6 } { \left( 1 - u ^ { 2 } \right) ( 2 - u ) }\) in partial fractions.
(iii) Hence, evaluate $$\int _ { 0 } ^ { \frac { \pi } { 6 } } \frac { 6 } { \cos x ( 2 - \sin x ) } d x$$ giving your answer in the form \(a \ln 2 + b \ln 3\), where \(a\) and \(b\) are integers.
OCR C4 2009 January Q5
8 marks Standard +0.3
5
  1. Show that the substitution \(u = \sqrt { x }\) transforms \(\int \frac { 1 } { x ( 1 + \sqrt { x } ) } \mathrm { d } x\) to \(\int \frac { 2 } { u ( 1 + u ) } \mathrm { d } u\).
  2. Hence find the exact value of \(\int _ { 1 } ^ { 9 } \frac { 1 } { x ( 1 + \sqrt { x } ) } \mathrm { d } x\).
OCR H240/01 2023 June Q12
10 marks Challenging +1.2
12
  1. Use the substitution \(u = \mathrm { e } ^ { x } - 2\) to show that $$\int \frac { 7 \mathrm { e } ^ { x } - 8 } { \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } } \mathrm {~d} x = \int \frac { 7 u + 6 } { u ^ { 2 } ( u + 2 ) } \mathrm { d } u$$
  2. Hence show that $$\int _ { \ln 4 } ^ { \ln 6 } \frac { 7 \mathrm { e } ^ { x } - 8 } { \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } } \mathrm {~d} x = a + \ln b$$ where \(a\) and \(b\) are rational numbers to be determined. \section*{END OF QUESTION PAPER}
Edexcel C4 Q8
15 marks Standard +0.3
8. (a) Show that the substitution \(u = \sin x\) transforms the integral $$\int \frac { 6 } { \cos x ( 2 - \sin x ) } d x$$ into the integral $$\int \frac { 6 } { \left( 1 - u ^ { 2 } \right) ( 2 - u ) } \mathrm { d } u .$$ (b) Express \(\frac { 6 } { \left( 1 - u ^ { 2 } \right) ( 2 - u ) }\) in partial fractions.
(c) Hence, evaluate $$\int _ { 0 } ^ { \frac { \pi } { 6 } } \frac { 6 } { \cos x ( 2 - \sin x ) } d x$$ giving your answer in the form \(a \ln 2 + b \ln 3\), where \(a\) and \(b\) are integers.
8. continued
8. continued