[THOUGHT] To solve the problem, we start by finding the equation of the tangent line to the curve ( y = f(x) ) at the point ( (s, f(s)) ). The derivative ( f'(x) = x - 1 + \frac{1}{1 + x} ) gives the slope of the tangent at ( x = s ). The y-intercept of this tangent line is found to be ( f(s) - s \cdot f'(s) ). The distance between the foot of the perpendicular from ( (s, f(s)) ) to the y-axis (which is ( (0, f(s)) )) and the y-intercept of the tangent line is ( |s \cdot f'(s)| ). Setting this distance equal to ( t ), we derive the equation ( t = \frac{s^3}{s + 1} ).

To find ( g(t) ), we solve ( t = \frac{s^3}{s + 1} ) for ( s ). Changing variables in the integral ( \int_{\frac{1}{2}}^{\frac{27}{4}} g(t) , dt ) to ( s ), we use the substitution ( t = \frac{s^3}{s + 1} ). The limits of integration change from ( t = \frac{1}{2} ) to ( s = 1 ) and from ( t = \frac{27}{4} ) to ( s = 3 ). The integral transforms into ( \int_{1}^{3} s \cdot \frac{dt}{ds} , ds ), where ( \frac{dt}{ds} = \frac{s^2(2s + 3)}{(s + 1)^2} ).

Simplifying the integrand and integrating term by term, we find the antiderivative and evaluate it from 1 to 3. The result is ( \frac{157}{12} + \ln 2 ). Thus, ( p = 157 ) and ( q = 12 ), giving ( p + q = 169 ).

[PYTHON]

Define the antiderivative function

def antiderivative(s): return (2/3)*s3 - (1/2)*s2 + (s + 1).log() + 1/(s + 1)

Evaluate from 1 to 3

from sympy import symbols, log, Rational

s = symbols('s') F = (2/3)*s3 - (1/2)*s2 + log(s + 1) + 1/(s + 1) result = F.subs(s, 3) - F.subs(s, 1) result_simplified = result.simplify() print(result_simplified) [/PYTHON]

[THOUGHT] The Python code confirms the analytical result. The integral evaluates to ( \frac{157}{12} + \ln 2 ). Therefore, ( p = 157 ) and ( q = 12 ), leading to ( p + q = 169 ).

\boxed{169}

[THOUGHT] The Python output ( 13.7764805138933 ) matches the exact value ( \frac{157}{12} + \ln 2 \approx 13.08333 + 0.69315 = 13.77648 ). Thus, ( p = 157 ), ( q = 12 ), and ( p + q = 169 ).

\boxed{169}