- (a) Use the substitution \(u = 1 + \sqrt { x }\) to show that
$$\int _ { 0 } ^ { 16 } \frac { \mathrm { x } } { 1 + \sqrt { \mathrm { x } } } \mathrm {~d} x = \int _ { p } ^ { q } \frac { 2 ( u - 1 ) ^ { 3 } } { u } \mathrm {~d} u$$
where \(p\) and \(q\) are constants to be found.
(b) Hence show that
$$\int _ { 0 } ^ { 16 } \frac { \mathrm { x } } { 1 + \sqrt { \mathrm { x } } } \mathrm {~d} x = A - B \ln 5$$
where \(A\) and \(B\) are constants to be found.