Use the substitution \(u ^ { 2 } = x ^ { 2 } + 3\) to show that \(\int \frac { 4 x ^ { 3 } } { \sqrt { x ^ { 2 } + 3 } } \mathrm {~d} x = \frac { 4 } { 3 } \left( x ^ { 2 } - 6 \right) \sqrt { x ^ { 2 } + 3 } + c\).
In this question you must show detailed reasoning.
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The graph shows part of the curve \(y = \frac { 4 x ^ { 3 } } { \sqrt { x ^ { 2 } + 2 } }\).
Find the exact area enclosed by the curve \(y = \frac { 4 x ^ { 3 } } { \sqrt { x ^ { 2 } + 3 } }\), the normal to this curve at the point \(( 1,2 )\) and the \(x\)-axis.