6. (a) Use the substitution \(u = \sqrt { t }\) to show that $$\int _ { 1 } ^ { x } \frac { \ln t } { \sqrt { t } } \mathrm {~d} t = 4 - 4 \sqrt { x } + 2 \sqrt { x } \ln x \quad x \geqslant 1$$ (b) The function g is such that $$\int _ { 1 } ^ { x } \mathrm {~g} ( t ) \mathrm { d } t = x - \sqrt { x } \ln x - 1 \quad x \geqslant 1$$

1. Use differentiation to find the function g .
2. Evaluate \(\int _ { 4 } ^ { 16 } \mathrm {~g} ( t ) \mathrm { d } t\) and simplify your answer.
   (c) Find the value of \(x\) (where \(x > 1\) ) that gives the maximum value of $$\int _ { x } ^ { x + 1 } \frac { \ln t } { 2 ^ { t } } \mathrm {~d} t$$