[THOUGHT] Let me analyze this problem step by step.

1. We have $f(x) = \frac{1}{2}x^2 - x + \ln(1+x)$
2. For a point $(s, f(s))$ on the curve:
   • The foot of the perpendicular to the y-axis is $(0, f(s))$
   • The tangent line at $(s, f(s))$ has slope $f'(s)$
   • We need to find where this tangent meets the y-axis

First, let me find $f'(x)$: $f'(x) = x - 1 + \frac{1}{1+x}$

The tangent line at $(s, f(s))$ is: $y - f(s) = f'(s)(x - s)$

At $x = 0$: $y = f(s) - s \cdot f'(s)$

So the tangent meets the y-axis at $(0, f(s) - s \cdot f'(s))$.

The distance between $(0, f(s))$ and $(0, f(s) - s \cdot f'(s))$ is: $|s \cdot f'(s)| = t$

Since $s > 0$, we need $|f'(s)| = t/s$, so $s \cdot |f'(s)| = t$.

Let me compute $f'(s) = s - 1 + \frac{1}{1+s} = \frac{(s-1)(1+s) + 1}{1+s} = \frac{s^2 - 1 + 1}{1+s} = \frac{s^2}{1+s}$

So $s \cdot f'(s) = s \cdot \frac{s^2}{1+s} = \frac{s^3}{1+s}$

For $s > 0$, this is always positive, so: $t = \frac{s^3}{1+s}$

We need to find $g(t)$ which is the value of $s$ such that $t = \frac{s^3}{1+s}$.

So $g(t)$ is the inverse function: given $t$, find $s$ such that $\frac{s^3}{1+s} = t$.

Now we need to compute $\int_{1/2}^{27/4} g(t) \, dt$.

Using substitution: let $t = \frac{s^3}{1+s}$, then $dt = \frac{d}{ds}\left(\frac{s^3}{1+s}\right) ds$

$\frac{d}{ds}\left(\frac{s^3}{1+s}\right) = \frac{3s^2(1+s) - s^3}{(1+s)^2} = \frac{3s^2 + 3s^3 - s^3}{(1+s)^2} = \frac{3s^2 + 2s^3}{(1+s)^2} = \frac{s^2(3 + 2s)}{(1+s)^2}$

When $t = 1/2$: $\frac{s^3}{1+s} = \frac{1}{2}$, so $2s^3 = 1 + s$, giving $2s^3 - s - 1 = 0$. When $t = 27/4$: $\frac{s^3}{1+s} = \frac{27}{4}$, so $4s^3 = 27 + 27s$, giving $4s^3 - 27s - 27 = 0$.

Let me solve these and compute the integral. [/THOUGHT] [PYTHON] from sympy import *

s = symbols('s', positive=True, real=True) t = symbols('t', positive=True, real=True)

f(x) = (1/2)x^2 - x + ln(1+x)

f = Rational(1,2)*s**2 - s + ln(1+s) f_prime = diff(f, s) print("f'(s) =", f_prime) print("Simplified f'(s) =", simplify(f_prime))

Distance = s * f'(s)

distance = s * f_prime distance_simplified = simplify(distance) print("s * f'(s) =", distance_simplified)

So t = s^3/(1+s), and g(t) is the inverse

We need to find the bounds for s

When t = 1/2: s^3/(1+s) = 1/2

eq1 = Eq(s**3/(1+s), Rational(1,2)) s_lower = solve(eq1, s) print("When t = 1/2, s =", s_lower)

When t = 27/4: s^3/(1+s) = 27/4

eq2 = Eq(s**3/(1+s), Rational(27,4)) s_upper = solve(eq2, s) print("When t = 27/4, s =", s_upper) [/PYTHON]

[THOUGHT] Now I can compute the integral using substitution. We have:

• $t = \frac{s^3}{1+s}$
• When $t = 1/2$, $s = 1$
• When $t = 27/4$, $s = 3$

Using substitution with $t = \frac{s^3}{1+s}$: $dt = \frac{s^2(3 + 2s)}{(1+s)^2} ds$

$\int_{1/2}^{27/4} g(t) \, dt = \int_1^3 s \cdot \frac{s^2(3 + 2s)}{(1+s)^2} ds = \int_1^3 \frac{s^3(3 + 2s)}{(1+s)^2} ds$

Let me compute this integral: $\int_1^3 \frac{s^3(3 + 2s)}{(1+s)^2} ds = \int_1^3 \frac{3s^3 + 2s^4}{(1+s)^2} ds$

I need to perform polynomial division or partial fractions. Let me compute this directly.

Actually, let me verify with Python to get the exact answer. [/THOUGHT]

Wait, I need to run the code to compute the integral. Let me do that.

[THOUGHT] I need to compute the integral $\int_1^3 \frac{s^3(3 + 2s)}{(1+s)^2} ds$ using Python. [/THOUGHT] [PYTHON] from sympy import *

s = symbols('s', positive=True, real=True)

The integrand after substitution

integrand = s**3 * (3 + 2*s) / (1+s)**2 integrand_simplified = simplify(integrand) print("Integrand:", integrand_simplified)

Compute the integral from 1 to 3

result = integrate(integrand, (s, 1, 3)) result_simplified = simplify(result) print("Integral result:", result_simplified)

Express in form p/q + ln(2)

Let's expand and see

result_expanded = expand(result_simplified) print("Expanded:", result_expanded)

Separate rational and logarithmic parts

result_rational = result_simplified.as_independent(ln(2))[0] result_log_coeff = result_simplified.coeff(ln(2)) print("Rational part:", result_rational) print("Coefficient of ln(2):", result_log_coeff)

If the form is p/q + ln(2), then coefficient of ln(2) should be 1

Let's rewrite

from sympy import apart print("\nAlternative approach:") print("Result =", result_simplified)

Check if it's in form a + b*ln(2)

a, b = result_simplified.as_independent(ln(2)) print("a =", a, ", b =", b) [/PYTHON]

[THOUGHT] The integral evaluates to $\frac{157}{12} + \ln 2$.

So we have $\frac{p}{q} + \ln 2 = \frac{157}{12} + \ln 2$, which means $p = 157$ and $q = 12$.

Therefore, $p + q = 157 + 12 = 169$. [/THOUGHT]

\boxed{169}