7 Let \(I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 4 } \right) ^ { n } } \mathrm {~d} x\). By considering \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \frac { x } { \left( 1 + x ^ { 4 } \right) ^ { n } } \right)\), show that
$$4 n I _ { n + 1 } = \frac { 1 } { 2 ^ { n } } + ( 4 n - 1 ) I _ { n }$$
Given that \(I _ { 1 } = 0.86697\), correct to 5 decimal places, find \(I _ { 3 }\).