9. (i) Show that the two non-stationary points of inflection on the curve \(y = \ln \left( 1 + 4 x ^ { 2 } \right)\) are at \(x = \pm \frac { 1 } { 2 }\).
\includegraphics[max width=\textwidth, alt={}, center]{f9e0bca6-c2a3-4868-b38b-942ceabd4992-20_492_1064_237_513} The diagram shows the curve \(y = \ln \left( 1 + 4 x ^ { 2 } \right)\). The shaded region is bounded by the curve and a line parallel to the \(x\)-axis which meets the curve where \(x = \frac { 1 } { 2 }\) and \(x = - \frac { 1 } { 2 }\).
(ii) Show that the area of the shaded region is given by $$\int _ { 0 } ^ { \ln 2 } \sqrt { \mathrm { e } ^ { y } - 1 } \mathrm {~d} y$$ (iii) Show that the substitution \(\mathrm { e } ^ { y } = \sec ^ { 2 } \theta\) transforms the integral in part (ii) to \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } 2 \tan ^ { 2 } \theta \mathrm {~d} \theta\).
(iv) Hence find the exact area of the shaded region.
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