[THOUGHT] The problem asks to evaluate the integral $\int_{1/2}^{27/4} g(t) \, dt$, where $g(t)$ is the inverse function of $t(s) = \frac{s^3}{1+s}$. Let $s = g(t)$, so $t = \frac{s^3}{1+s}$. The integral becomes $\int_{s_1}^{s_2} s \frac{dt}{ds} \, ds$. First, I need to find the limits of integration $s_1$ and $s_2$ corresponding to $t = 1/2$ and $t = 27/4$. Then I will compute the derivative $\frac{dt}{ds}$ and the integral.

Steps:

1. Solve $\frac{s^3}{1+s} = \frac{1}{2}$ for $s > 0$ to find $s_1$.
2. Solve $\frac{s^3}{1+s} = \frac{27}{4}$ for $s > 0$ to find $s_2$.
3. Define the integrand $s \cdot \frac{d}{ds}\left(\frac{s^3}{1+s}\right)$.
4. Compute the definite integral using SymPy.
5. Format the result as $\frac{p}{q} + \ln 2$ and find $p+q$. [/THOUGHT] [PYTHON] from sympy import symbols, solve, diff, integrate, ln, simplify, Rational

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

Define t as a function of s

t_expr = s**3 / (1 + s)

Find s1 for t = 1/2

s^3 / (1+s) = 1/2 => 2s^3 - s - 1 = 0

s1_solutions = solve(2*s**3 - s - 1, s) print(f"s1 solutions: {s1_solutions}")

Find s2 for t = 27/4

s^3 / (1+s) = 27/4 => 4s^3 - 27s - 27 = 0

s2_solutions = solve(4s**3 - 27s - 27, s) print(f"s2 solutions: {s2_solutions}")

Calculate dt/ds

dt_ds = diff(t_expr, s) print(f"dt/ds: {dt_ds}")

Integrand is s * dt/ds

integrand = s * dt_ds print(f"Integrand: {integrand}")

We need to pick the correct real positive roots

Let's filter them in the next step or manually inspect output

[/PYTHON] [THOUGHT] The previous code will output the solutions. I need to identify the positive real roots $s_1$ and $s_2$. Then I will perform the integration.

Let's anticipate the roots: For $2s^3 - s - 1 = 0$: $s=1$ works ($2-1-1=0$). So $s_1 = 1$. For $4s^3 - 27s - 27 = 0$: Let's test small integers or fractions. $s=3$: $4(27) - 27(3) - 27 = 108 - 81 - 27 = 0$. So $s_2 = 3$.

So the limits are $s_1 = 1$ and $s_2 = 3$. I will verify this in the code and then compute the integral $\int_{1}^{3} s \frac{d}{ds}\left(\frac{s^3}{1+s}\right) ds$. The integral expression is $\int_{1}^{3} \frac{s^2(2s+3)}{(1+s)^2} ds$. I will use SymPy to compute this exact value. [/THOUGHT] [PYTHON] from sympy import symbols, integrate, simplify, ln

s = symbols('s', real=True, positive=True) t_expr = s**3 / (1 + s) dt_ds = t_expr.diff(s) integrand = s * dt_ds

Limits

s1 = 1 s2 = 3

Compute integral

result = integrate(integrand, (s, s1, s2)) print(f"Result: {result}") print(f"Simplified Result: {simplify(result)}") [/PYTHON] [THOUGHT] The output of the previous code will give me the value of the integral. Let's assume the result is something like A + ln(2). I need to match it to $\frac{p}{q} + \ln 2$. Then compute $p+q$.

Let's double check the manual integration plan to be sure. Integrand: $s \cdot \frac{3s^2(1+s) - s^3}{(1+s)^2} = \frac{s(3s^2 + 3s^3 - s^3)}{(1+s)^2} = \frac{2s^4 + 3s^3}{(1+s)^2}$. Wait, my previous manual derivative was $\frac{2s^3+3s^2}{(1+s)^2}$. Let's recompute $\frac{dt}{ds}$. $t = \frac{s^3}{1+s}$. $t' = \frac{3s^2(1+s) - s^3(1)}{(1+s)^2} = \frac{3s^2 + 3s^3 - s^3}{(1+s)^2} = \frac{2s^3 + 3s^2}{(1+s)^2}$. Correct. Then multiply by $s$ (since we are integrating $s \, dt = s \frac{dt}{ds} ds$). So integrand is $s \cdot \frac{2s^3 + 3s^2}{(1+s)^2} = \frac{2s^4 + 3s^3}{(1+s)^2}$.

Polynomial division: Numerator: $2s^4 + 3s^3$. Denominator: $s^2 + 2s + 1$. $2s^4 + 3s^3 = 2s^2(s^2 + 2s + 1) - 4s^3 - 2s^2 + 3s^3 = 2s^2(s+1)^2 - s^3 - 2s^2$. This seems tedious. Let $u = 1+s$, so $s = u-1$. Integrand = $\frac{2(u-1)^4 + 3(u-1)^3}{u^2}$. $= \frac{(u-1)^3 [2(u-1) + 3]}{u^2} = \frac{(u-1)^3 (2u + 1)}{u^2}$. Expand numerator: $(u^3 - 3u^2 + 3u - 1)(2u + 1)$ $= 2u^4 + u^3 - 6u^3 - 3u^2 + 6u^2 + 3u - 2u - 1$ $= 2u^4 - 5u^3 + 3u^2 + u - 1$. Divide by $u^2$: $2u^2 - 5u + 3 + \frac{1}{u} - \frac{1}{u^2}$. Integrate term by term with respect to $u$. Limits for $s$: 1 to 3. Limits for $u$: 2 to 4. $\int_2^4 (2u^2 - 5u + 3 + u^{-1} - u^{-2}) du$ $= [\frac{2}{3}u^3 - \frac{5}{2}u^2 + 3u + \ln|u| + u^{-1}]_2^4$.

Evaluate at 4: $\frac{2}{3}(64) - \frac{5}{2}(16) + 3(4) + \ln 4 + \frac{1}{4}$ $= \frac{128}{3} - 40 + 12 + 2\ln 2 + 0.25$ $= 42.66... - 28 + ...$ $= \frac{128}{3} - 28 + 0.25 + 2\ln 2$ $= \frac{128 - 84}{3} + \frac{1}{4} + 2\ln 2$ $= \frac{44}{3} + \frac{1}{4} + 2\ln 2$ $= \frac{176 + 3}{12} + 2\ln 2 = \frac{179}{12} + 2\ln 2$.

Evaluate at 2: $\frac{2}{3}(8) - \frac{5}{2}(4) + 3(2) + \ln 2 + \frac{1}{2}$ $= \frac{16}{3} - 10 + 6 + \ln 2 + 0.5$ $= \frac{16}{3} - 4 + 0.5 + \ln 2$ $= \frac{16 - 12}{3} + \frac{1}{2} + \ln 2$ $= \frac{4}{3} + \frac{1}{2} + \ln 2$ $= \frac{8+3}{6} + \ln 2 = \frac{11}{6} + \ln 2$.

Difference: $(\frac{179}{12} + 2\ln 2) - (\frac{11}{6} + \ln 2)$ $= \frac{179}{12} - \frac{22}{12} + \ln 2$ $= \frac{157}{12} + \ln 2$.

So $p = 157, q = 12$. $p+q = 169$.

Let's verify this with the Python code. [/THOUGHT] [PYTHON] from sympy import symbols, integrate, simplify, ln, Rational

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

Integrand derived: (2s^4 + 3s^3) / (1+s)^2

integrand = (2s**4 + 3s**3) / (1 + s)**2

Limits

s1 = 1 s2 = 3

Compute integral

result = integrate(integrand, (s, s1, s2)) print(f"Result: {result}")

Check if result matches 157/12 + ln(2)

expected = Rational(157, 12) + ln(2) print(f"Expected: {expected}") print(f"Difference: {simplify(result - expected)}") [/PYTHON] [THOUGHT] The code will confirm the result. Result should be 157/12 + log(2). The question asks for $p+q$ where the value is $\frac{p}{q} + \ln 2$. $p = 157$, $q = 12$. $p+q = 157 + 12 = 169$.

Wait, I should double check if the fraction can be simplified. 157 is not divisible by 2 (odd). Sum of digits 1+5+7 = 13, not divisible by 3. 157 ends in 7, not 5. 157 / 7 = 22.4... 157 / 11 = 14.2... 157 / 13 = 12.07... Square root of 157 is approx 12.5. Primes to check: 2, 3, 5, 7, 11. None work. So 157 is likely prime. Thus the fraction is in simplest form.

The problem statement says "If the value is $\frac{p}{q} + \ln 2$". It does not explicitly say "in simplest form", but usually that is implied or $p, q$ are coprime integers. Given the specific form, I will assume $p=157, q=12$.

Let's double check the question wording carefully. "If the value is $\frac{p}{q} + \ln 2$, compute $p+q$." This implies the coefficient of $\ln 2$ is exactly 1. My manual calculation gave $\ln 2$ (from $2\ln 2 - \ln 2$). So the form matches.

Let's verify the integration steps again. $u = s+1$. Integrand in $u$: $2u^2 - 5u + 3 + u^{-1} - u^{-2}$. Integral: $\frac{2}{3}u^3 - \frac{5}{2}u^2 + 3u + \ln u + u^{-1}$. Upper limit $u=4$: $2/3(64) - 5/2(16) + 12 + \ln 4 + 1/4$ $= 128/3 - 40 + 12 + 2\ln 2 + 1/4$ $= 128/3 - 28 + 1/4 + 2\ln 2$ $= (512 - 336 + 3)/12 + 2\ln 2$ $= 179/12 + 2\ln 2$. Correct.

Lower limit $u=2$: $2/3(8) - 5/2(4) + 6 + \ln 2 + 1/2$ $= 16/3 - 10 + 6 + \ln 2 + 1/2$ $= 16/3 - 4 + 1/2 + \ln 2$ $= (32 - 24 + 3)/6 + \ln 2$ $= 11/6 + \ln 2$. Correct.

Difference: $179/12 - 22/12 + (2\ln 2 - \ln 2)$ $= 157/12 + \ln 2$. Correct.

Final check of the problem statement details. $f(x) = \frac{1}{2}x^2 - x + \ln(1+x)$. $t = s f'(s)$. $f'(x) = x - 1 + \frac{1}{1+x} = \frac{x^2}{1+x}$. $t = \frac{s^3}{1+s}$. Integral limits: $t=1/2$ to $t=27/4$. $s_1$: $2s^3 - s - 1 = 0 \implies s=1$. $s_2$: $4s^3 - 27s - 27 = 0 \implies s=3$. Integral is $\int_{1/2}^{27/4} g(t) dt = \int_1^3 s \frac{dt}{ds} ds$. Everything looks consistent.

One detail: "distance between the foot of the perpendicular ... and the point where the tangent ... meets the y-axis". Foot of perp from $(s, f(s))$ to y-axis is $(0, f(s))$. Tangent y-intercept is $(0, f(s) - s f'(s))$. Distance is $|f(s) - (f(s) - s f'(s))| = |s f'(s)|$. Since $s>0$ and $f'(s) = \frac{s^2}{1+s} > 0$, distance is $s f'(s)$. This matches my derivation.

The integral is $\int g(t) dt$. Let $u = g(t)$, so $t = f(u)$ (using $f$ for the relation between $t$ and $s$, not the original function). Actually, let's stick to $s$ and $t$. $s = g(t)$. $\int_{t_1}^{t_2} s \, dt$. Integration by parts? $\int s \, dt = [st] - \int t \, ds$. Let's check this alternative method. $[st]_{s=1}^{s=3} - \int_1^3 \frac{s^3}{1+s} ds$. $t(3) = \frac{27}{4}$. $s=3 \implies st = 3 \cdot \frac{27}{4} = \frac{81}{4}$. $t(1) = \frac{1}{2}$. $s=1 \implies st = 1 \cdot \frac{1}{2} = \frac{1}{2}$. Boundary term: $\frac{81}{4} - \frac{1}{2} = \frac{81-2}{4} = \frac{79}{4}$.

Integral term: $\int_1^3 \frac{s^3}{1+s} ds$. $\frac{s^3}{1+s} = \frac{s^3+1-1}{1+s} = s^2 - s + 1 - \frac{1}{1+s}$. Integral is $[\frac{1}{3}s^3 - \frac{1}{2}s^2 + s - \ln(1+s)]_1^3$. At 3: $9 - 4.5 + 3 - \ln 4 = 7.5 - 2\ln 2$. At 1: $1/3 - 1/2 + 1 - \ln 2 = 5/6 - \ln 2$. Difference: $(7.5 - 5/6) + (-2\ln 2 + \ln 2)$. $7.5 = 15/2 = 45/6$. $45/6 - 5/6 = 40/6 = 20/3$. So integral part is $20/3 - \ln 2$.

Total result: $\frac{79}{4} - (\frac{20}{3} - \ln 2)$. $= \frac{79}{4} - \frac{20}{3} + \ln 2$. $= \frac{237 - 80}{12} + \ln 2$ $= \frac{157}{12} + \ln 2$.

Matches perfectly. $p=157, q=12$. $p+q = 169$.

One final check on the code execution to be absolutely sure. [/THOUGHT] [PYTHON] from sympy import symbols, integrate, ln, Rational

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

Method 1: Direct integration

integrand = (2s**4 + 3s**3) / (1 + s)**2 res1 = integrate(integrand, (s, 1, 3)) print(f"Method 1: {res1}")

Method 2: Integration by parts

int s dt = [st] - int t ds

term1 = Rational(79, 4) term2 = integrate(s**3 / (1 + s), (s, 1, 3)) res2 = term1 - term2 print(f"Method 2: {res2}")

Check equality

print(f"Equal? {res1 == res2}") [/PYTHON] [THOUGHT] Both methods yield the same result. The value is $\frac{157}{12} + \ln 2$. $p = 157$. $q = 12$. $p+q = 169$.

The question asks for $p+q$. Final Answer seems to be 169. [/THOUGHT] [THOUGHT] I have verified the solution using two different integration methods and Python code. The value of the integral is $\frac{157}{12} + \ln 2$. This is in the form $\frac{p}{q} + \ln 2$. $p = 157$, $q = 12$. $p+q = 157 + 12 = 169$. I am confident in this result. [/THOUGHT] $\boxed{169}$

$\boxed{169}$