Partial fractions after substitution

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.

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 $$\int \frac { 1 } { 4 x + 5 x ^ { \frac { 1 } { 3 } } } \mathrm {~d} x , \quad x > 0$$

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.)

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 \(\int _ { 1 } ^ { 9 } \frac { 1 } { x ( 1 + \sqrt { x } ) } \mathrm { d } x\).

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}

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

Using partial fractions, find \(\int \frac{x}{(x+1)(2x+1)} \, dx\). [7]

Using partial fractions, find \(\int \frac{x}{(x+1)(2x+1)} dx\). [7]