Show definite integral equals specific value (requiring partial fractions or complex algebra)

7 Let \(I = \int _ { 1 } ^ { 4 } \frac { 1 } { x ( 4 - \sqrt { } x ) } \mathrm { d } x\).

1. Use the substitution \(u = \sqrt { } x\) to show that \(I = \int _ { 1 } ^ { 2 } \frac { 2 } { u ( 4 - u ) } \mathrm { d } u\).
2. Hence show that \(I = \frac { 1 } { 2 } \ln 3\).

8 Let \(I = \int _ { 2 } ^ { 5 } \frac { 5 } { x + \sqrt { } ( 6 - x ) } \mathrm { d } x\).

1. Using the substitution \(u = \sqrt { } ( 6 - x )\), show that $$I = \int _ { 1 } ^ { 2 } \frac { 10 u } { ( 3 - u ) ( 2 + u ) } \mathrm { d } u$$
2. Hence show that \(I = 2 \ln \left( \frac { 9 } { 2 } \right)\).

10 By first using the substitution \(u = \mathrm { e } ^ { x }\), show that $$\int _ { 0 } ^ { \ln 4 } \frac { \mathrm { e } ^ { 2 x } } { \mathrm { e } ^ { 2 x } + 3 \mathrm { e } ^ { x } + 2 } \mathrm {~d} x = \ln \left( \frac { 8 } { 5 } \right)$$

6 Let \(I = \int _ { 1 } ^ { 4 } \frac { ( \sqrt { } x ) - 1 } { 2 ( x + \sqrt { } x ) } \mathrm { d } x\).

1. Using the substitution \(u = \sqrt { } x\), show that \(I = \int _ { 1 } ^ { 2 } \frac { u - 1 } { u + 1 } \mathrm {~d} u\).
2. Hence show that \(I = 1 + \ln \frac { 4 } { 9 }\).

6. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation $$y = \frac { 16 \sin 2 x } { ( 3 + 4 \sin x ) ^ { 2 } } \quad 0 \leqslant x \leqslant \frac { \pi } { 2 }$$ The region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the line with equation \(x = \frac { \pi } { 6 }\) Using the substitution \(u = 3 + 4 \sin x\), show that the area of \(R\) can be written in the form \(a + \ln b\), where \(a\) and \(b\) are rational constants to be found.

4. Use the substitution \(x + 2 = u ^ { 2 }\), where \(u > 0\), to show that $$\int _ { - 1 } ^ { 7 } \frac { ( x + 2 ) ^ { \frac { 1 } { 2 } } } { x + 5 } \mathrm {~d} x = a + b \pi \sqrt { 3 }$$ where \(a\) and \(b\) are rational numbers to be found. \includegraphics[max width=\textwidth, alt={}, center]{image-not-found}
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1. (a) Use the substitution \(x = u ^ { 2 } + 1\) to show that

$$\int _ { 5 } ^ { 10 } \frac { 3 \mathrm {~d} x } { ( x - 1 ) ( 3 + 2 \sqrt { x - 1 } ) } = \int _ { p } ^ { q } \frac { 6 \mathrm {~d} u } { u ( 3 + 2 u ) }$$ where \(p\) and \(q\) are positive constants to be found.
(b) Hence, using algebraic integration, show that $$\int _ { 5 } ^ { 10 } \frac { 3 \mathrm {~d} x } { ( x - 1 ) ( 3 + 2 \sqrt { x - 1 } ) } = \ln a$$ where \(a\) is a rational constant to be found.