[THOUGHT] We need to evaluate the definite integral of g(t) from t = 1/2 to t = 27/4, where g(t) is the positive s such that t = s^3/(1+s). The integral I = ∫{1/2}^{27/4} g(t) dt. Using change of variable s = g(t), we have t = s^3/(1+s), dt = h'(s) ds with h(s) = s^3/(1+s). Then I = ∫{s=1}^{3} s h'(s) ds. Integration by parts gives I = [s h(s)]{1}^{3} - ∫{1}^{3} h(s) ds.

We compute: h(s) = s^3/(1+s) = s^2 - s + 1 - 1/(1+s). Thus s h(s) = s^4/(1+s) = s^3 - s^2 + s - 1 + 1/(1+s).

Evaluating from 1 to 3: s h(s) at 3: 27 - 9 + 3 - 1 + 1/4 = 20 + 1/4 = 81/4? Wait 27-9=18, 18+3=21, 21-1=20, plus 1/4 = 81/4? Actually 20 = 80/4, plus 1/4 = 81/4. So value = 81/4. s h(s) at 1: 1 - 1 + 1 - 1 + 1/2 = 1/2. Thus [s h(s)]_1^3 = 81/4 - 1/2 = 81/4 - 2/4 = 79/4? Wait 81/4 - 2/4 = 79/4? But earlier we got 77/4. Let's recalc: 81/4 - 1/2 = 81/4 - 2/4 = 79/4. But earlier we computed 77/4. Something is off. Let's double-check the expression for s h(s) at s=3 and s=1 using the formula s h(s) = s^4/(1+s). Compute directly: s=3: s^4 = 81, 1+s=4, so s h(s) = 81/4 = 20.25. At s=1: 1/(2) = 0.5. So difference is 20.25 - 0.5 = 19.75 = 79/4? Actually 20.25 - 0.5 = 19.75, which is 79/4? 79/4 = 19.75. Yes, 79/4 = 19.75. So [s h(s)]_1^3 = 81/4 - 1/2 = 81/4 - 2/4 = 79/4. Wait 1/2 = 2/4, so 81/4 - 2/4 = 79/4. But earlier I computed 77/4. That's a discrepancy. Let's recalc s h(s) at s h(s) using the polynomial expression: s h(s) = s^4/(1+s). Evaluate at s=3: 81/4 = 20.25. At s=1: 1/2 = 0.5. So difference = 19.75 = 79/4. So indeed [s h(s)]_1^3 = 79/4, not 77/4. But earlier we derived s h(s) = s^3 - s^2 + s - 1 + 1/(s+1). Let's check that: s h(s) = s * (s^2 - s + 1 - 1/(1+s))? Wait h(s) = s^2 - s + 1 - 1/(1+s). Then s h(s) = s^3 - s^2 + s - 1 + 1/(1+s)? Because s * (-1/(1+s)) = -s/(1+s). But we previously wrote s h(s) = s^3 - s^2 + s - 1 + 1/(s+1). Let's verify: s h(s) = s*(s^2 - s + 1 - 1/(1+s)) = s^3 - s^2 + s - s/(1+s). Now s/(1+s) = 1 - 1/(1+s). Because s/(1+s) = (1+s-1)/(1+s) = 1 - 1/(1+s). So - s/(1+s) = -1 + 1/(1+s). Then s h(s) = s^3 - s^2 + s - 1 + 1/(1+s). Yes, that matches: s^3 - s^2 + s - 1 + 1/(s+1). So our expression is correct.

Now evaluate at s=3: s^3 - s^2 + s - 1 + 1/(s+1) = 27 - 9 + 3 - 1 + 1/4 = (27-9=18, 18+3=21, 21-1=20) + 1/4 = 20 + 1/4 = 80/4 + 1/4? Wait 20 = 80/4, plus 1/4 = 81/4. So that gives 81/4. Good. At s=1: 1 - 1 + 1 - 1 + 1/2 = 0 + 0 + 1/2? Actually 1-1=0, 0+1=1, 1-1=0, +1/2 = 1/2. So s h(s) = 1/2. So difference = 81/4 - 1/2 = 81/4 - 2/4 = 79/4. So [s h(s)]_1^3 = 79/4, not 77/4. My earlier miscalculation: I had written 77/4, which is wrong. Let's recalc: 81/4 - 1/2 = (81 - 2)/4 = 79/4. So 79/4 is correct.

Thus I = 79/4 - ∫_1^3 h(s) ds.

Now compute ∫_1^3 h(s) ds. We have h(s) = s^2 - s + 1 - 1/(1+s). So ∫ h(s) ds = s^3/3 - s^2/2 + s - ln(1+s). Evaluate from 1 to 3.

At s=3: (27/3) - (9/2) + 3 - ln4 = 27/3? Wait s^3/3 = 27/3 = 9? Actually 27/3 = 9? No, 27/3 = 9? 27/3 = 9? That's wrong: 27/3 = 9? 39 = 27, yes 27/3 = 9. So s^3/3 = 9. But careful: s^3/3 = 27/3 = 9? Wait 27/3 = 9? 27/3 = 9? No, 27 divided by 3 is 9, yes because 39 = 27. So 9. But earlier we wrote 27/3 = 9, correct. But earlier we used 15/2 = 7.5, not 9. Let's compute properly:

s^3/3 at s=3: 3^3/3 = 27/3 = 9.

• s^2/2 = -9/2 = -4.5.

• s = +3. Sum = 9 - 4.5 + 3 = 7.5 = 15/2. So s^3/3 - s^2/2 + s = 15/2? Wait 9 - 4.5 = 4.5, +3 = 7.5 = 15/2. Yes, that's correct. So the polynomial part is 15/2, not 9 - 4.5 + 3 = 7.5. So that's 15/2. So the antiderivative F(s) = s^3/3 - s^2/2 + s - ln(1+s). So at s=3: F(3) = 27/3 - 9/2 + 3 - ln4 = 9 - 4.5 + 3 - ln4 = 7.5 - ln4 = 15/2 - ln4. At s=1: F(1) = 1/3 - 1/2 + 1 - ln2 = (2/6 - 3/6 + 6/6) = 5/6 - ln2. So ∫_1^3 h(s) ds = F(3) - F(1) = (15/2 - ln4) - (5/6 - ln2) = 15/2 - ln4 - 5/6 + ln2 = (45/6 - 5/6) + (-ln4 + ln2) = (40/6) + (-ln(4/2))? Actually -ln4 + ln2 = ln(2/4) = ln(1/2) = -ln2. So = 40/6 - ln2 = 20/3 - ln2.

Thus ∫_1^3 h(s) ds = 20/3 - ln2.

Therefore I = [s h(s)]_1^3 - ∫_1^3 h(s) ds = (79/4) - (20/3 - ln2) = 79/4 - 20/3 + ln2.

Now compute 79/4 - 20/3 = (793 - 204) / 12 = (237 - 80) / 12 = 157/12. Because 793 = 237, 204=80, 237-80=157. So I = 157/12 + ln2.

Thus the integral equals 157/12 + ln2.

But we must double-check the calculation of 79/4 - 20/3. 79/4 = 237/12, 20/3 = 80/12, difference = 157/12. Yes.

So I = 157/12 + ln2.

Now is that the integral from t=1/2 to 27/4? Wait, we had I = ∫_{1/2}^{27/4} g(t) dt = ∫_1^3 s h'(s) ds = [s h(s)]_1^3 - ∫_1^3 h(s) ds = 79/4 - (20/3 - ln2) = 79/4 - 20/3 + ln2 = (237/12 - 80/12) + ln2 = 157/12 + ln2.

So the integral is 157/12 + ln2.

But the problem statement says: "If the value is p/q + ln 2, compute p+q." So p/q = 157/12, with p=157, q=12, p+q = 169.

However, we must verify that our derivation of g(t) and the integration limits are correct. Let's re-derive carefully.

Given f(x) = (1/2)x^2 - x + ln(1+x). For a point (s, f(s)) on the curve. The foot of perpendicular to y-axis is (0, f(s)). The tangent line at (s, f(s)) has slope f'(s) = s^2/(1+s). Its y-intercept is at x=0: y = f(s) - f'(s) s. So the distance between (0, f(s)) and (0, f(s)- f'(s) s) is |f'(s) s| = f'(s) s, because f'(s)>0 for s>0. So t = f'(s) s = s * s^2/(1+s) = s^3/(1+s). So t = s^3/(1+s). Good.

Thus for each t>0, there is a unique s>0 such that t = s^3/(1+s). So g(t) = s.

Thus the integral I = ∫{t=1/2}^{27/4} g(t) dt = ∫{s: s^3/(1+s)=1/2}^{27/4} s * (dt/ds) ds? Actually dt = (d/ds)[s^3/(1+s)] ds. So I = ∫_{s=1}^{3} s * (d/ds)[s^3/(1+s)] ds = ∫_1^3 s h'(s) ds.

Yes.

Now we computed h(s) = s^3/(1+s). h'(s) = (3s^2(1+s) - s^3)/(1+s)^2 = (3s^2+3s^3 - s^3)/(1+s)^2 = (3s^2+2s^3)/(1+s)^2 = s^2(3+2s)/(1+s)^2.

But we didn't need h'(s) directly using integration by parts.

So I = ∫_1^3 s h'(s) ds = [s h(s)]_1^3 - ∫_1^3 h(s) ds.

Now we computed s h(s) at 3 = 81/4, at 1 = 1/2, so [s h(s)]_1^3 = 81/4 - 1/2 = 81/4 - 2/4 = 79/4. Good.

Now ∫_1^3 h(s) ds = ∫_1^3 s^3/(1+s) ds. We computed that integral as 20/3 - ln2. Let's verify with direct integration.

Compute ∫ s^3/(1+s) ds = ∫ (s^2 - s + 1 - 1/(1+s)) ds = s^3/3 - s^2/2 + s - ln|1+s| + C. Evaluate from 1 to 3:

At 3: 27/3 - 9/2 + 3 - ln4 = 9 - 4.5 + 3 - ln4 = 7.5 - ln4 = 15/2 - ln4. At 1: 1/3 - 1/2 + 1 - ln2 = (2/6 - 3/6 + 6/6) = 5/6 - ln2. Difference: (15/2 - ln4) - (5/6 - ln2) = 15/2 - 5/6 + (-ln4 + ln2) = (45/6 - 5/6) + (ln2 - ln4) = 40/6 + ln(2/4) = 20/3 + ln(1/2) = 20/3 - ln2. Yes, ∫ h(s) ds = 20/3 - ln2.

Thus I = 79/4 - (20/3 - ln2) = 79/4 - 20/3 + ln2 = (793 - 204) /12 + ln2 = (237 - 80)/12 + ln2 = 157/12 + ln2.

So I = 157/12 + ln2.

Thus p/q = 157/12, with p=157, q=12, relatively prime? 157 is prime? 157 is prime? Let's check divisibility: 2 no, 3: 1+5+7=13 not divisible by 3, 5 no, 7: 722=154, so no, 11: 1114=154, no, 13: 13*12=156, no. So 157 is prime. So p+q = 157+151? Wait q=12, not 152. Actually p+q = 157+12 = 169. But earlier we had 163? That was mistaken. So answer is 169.

But we must double-check the integration limits: t from 1/2 to 27/4. We found that at t=1/2, s=1; at t=27/4, s=3. That matches. So I = 157/12 + ln2.

Now, could there be any mistake? Let's double-check the integral of h(s) from 1 to 3: h(s) = s^3/(1+s). We computed the antiderivative as s^3/3 - s^2/2 + s - ln(1+s). But wait, is that the antiderivative? Because we had h(s) = s^2 - s + 1 - 1/(1+s). The integral of s^2 is s^3/3, integral of -s is -s^2/2, integral of +1 is +s, integral of -1/(1+s) is -ln|1+s|. So yes, that matches. So the antiderivative F(s) = s^3/3 - s^2/2 + s - ln(1+s). Evaluate at 3: 27/3 - 9/2 + 3 - ln4 = 9 - 4.5 + 3 - ln4 = 7.5 - ln4 = 15/2 - ln4. At 1: 1/3 - 1/2 + 1 - ln2 = (2/6 - 3/6 + 6/6) = 5/6 - ln2. So difference = (15/2 - ln4) - (5/6 - ln2) = (45/6 - 5/6) + (-ln4 + ln2) = 40/6 + ln(2/4) = 20/3 + ln(1/2) = 20/3 - ln2. So correct.

Thus I = 79/4 - (20/3 - ln2) = 79/4 - 20/3 + ln2 = (237/12 - 80/12) + ln2 = 157/12 + ln2.

So answer is 157/12 + ln2.

But wait, is the integral from t=1/2 to 27/4 indeed equal to ∫1^3 s h'(s) ds? Let's double-check the change of variable: t = h(s) = s^3/(1+s). Then dt = h'(s) ds. When t = 1/2, s = 1; when t = 27/4, s = 3. So I = ∫{t=1/2}^{27/4} g(t) dt = ∫_{s=1}^{3} s * (dt/ds) ds = ∫_1^3 s h'(s) ds. Yes.

Now, is there any nuance about absolute value? t is distance, so t>0. And we have t = s^3/(1+s) which is positive for s>0. So that's fine.

Thus I = 157/12 + ln2.

Now the problem says: "If the value is p/q + ln 2, compute p+q." So p/q = 157/12, so p=157, q=12, p+q = 169.

But we must ensure that p/q is in simplest terms? 157 and 12 are coprime, so p=157, q=12, sum=169.

Thus answer is 169.

But wait, check the problem statement: "If the value is p/q + ln 2, compute p+q." So they want p+q. So answer 169.

However, we should verify the initial function and the interpretation of "distance between the foot of the perpendicular from (s, f(s)) to the y-axis and the point where the tangent to the curve at (s, f(s)) meets the y-axis". The foot is (0, f(s)). The tangent line meets y-axis at (0, f(s) - s f'(s)). So distance = |f(s) - (f(s) - s f'(s))| = s f'(s). Yes.

So t = s f'(s). And f'(s) = s^2/(1+s). So t = s^3/(1+s). That's correct.

Now, is there any possibility that the foot of perpendicular is not (0, f(s))? The foot of perpendicular from a point to the y-axis: Actually "the foot of the perpendicular from (s, f(s)) to the y-axis" means the perpendicular from point to the y-axis. Since the y-axis is a line, the foot is the point on y-axis closest to (s, f(s)). That is indeed (0, f(s)). Because the line through (s, f(s)) perpendicular to y-axis is horizontal (parallel to x-axis), so it meets y-axis at (0, f(s)). So correct.

Thus everything consistent.

Now we must ensure the integration limits: t runs from 1/2 to 27/4. We solved s^3/(1+s) = 1/2 gave s=1. And =27/4 gave s=3. So that matches.

Thus final answer is 169.

But before finalizing, let's double-check the integration by parts: I = ∫_1^3 s h'(s) ds = [s h(s)]_1^3 - ∫_1^3 h(s) ds. That is correct.

Now compute [s h(s)]_1^3 directly: s h(s) = s * s^3/(1+s) = s^4/(1+s). At s=3: 81/4 = 20.25. At s=1: 1/2 = 0.5. So difference = 19.75 = 79/4. Yes.

Now ∫_1^3 h(s) ds = ∫_1^3 s^3/(1+s) ds. Compute numerically to check: s from 1 to 3, function s^3/(1+s). At s=1: 1/2=0.5; at s=2: 8/3≈2.6667; at s=3: 27/4=6.75. The integral is roughly average ~ maybe 3? Let's approximate: use Simpson? But we trust symbolic. Compute numerically: ∫_1^3 s^3/(1+s) ds. Let's do numeric integration quickly: Use antiderivative F(s)= s^3/3 - s^2/2 + s - ln(1+s). At s=3: 9 - 4.5 + 3 - ln4 = 7.5 - 1.3863 = 6.1137? Wait 7.5 - 1.3863 = 6.1137. At s=1: 0.3333 - 0.5 + 1 - ln2 = 0.8333 - 0.6931 = 0.1402? Actually compute: 0.3333 - 0.5 = -0.1667, +1 = 0.8333, minus ln2 (0.6931) = 0.1402. So integral = 6.1137 - 0.1402 = 5.9735? That seems off: 6.1137 - 0.1402 = 5.9735. But earlier we computed 20/3 - ln2 ≈ 6.6667 - 0.6931 = 5.9736. Yes, that matches. So ∫_1^3 h(s) ds ≈ 5.9736. Then I = 19.75 - 5.9736 = 13.7764? Wait 19.75 - 5.9736 = 13.7764. But we got I = 157/12 + ln2 ≈ 13.08333 + 0.6931 = 13.7764. Yes, 157/12 ≈ 13.08333, plus ln2 ≈ 0.693147 = 13.77644. So consistent.

Thus answer is 169.

But is there any subtlety about the distance being between the foot and the intersection point? It says "distance between the foot of the perpendicular from (s, f(s)) to the y-axis and the point where the tangent to the curve at (s, f(s)) meets the y-axis". The foot is on the y-axis, the intersection point is also on the y-axis, so distance is absolute difference of their y-coordinates. That's what we used.

Thus final answer: \boxed{169}.

But we must output the answer as \boxed{169}. However, the instruction: "If the value is p/q + ln 2, compute p+q." So we output p+q = 169. So final answer: 169.

But wait, check the problem statement: "If the value is p/q + ln 2, compute p+q." It might be that the value is something like (something) + ln2, and we need p+q. So we have p/q = 157/12, p=157, q=12, sum=169.

Thus answer is 169.

Now, we must ensure that the problem didn't expect the integral from t=1/2 to 27/4 of g(t) dt to equal p/q + ln 2, and we found it to be 157/12 + ln2. So p=157, q=12, p+q=169.

Double-check the integration limits: t=1/2 and t=27/4. Are we sure that the lower limit is 1/2? The problem says: ∫_{1/2}^{27/4} g(t) dt. Yes.

Now, we should verify the s value for t=1/2: solve s^3/(1+s) = 1/2 => 2s^3 = 1+ s => 2s^3 - s - 1 = 0. s=1 is a root: 2-1-1=0. Factor (s-1)(2s^2+2s+1)=0? Actually divide: (2s^3 - s -1) ÷ (s-1). Perform polynomial division: 2s^3 +0s^2 - s -1 divided by (s-1). Using synthetic: 1 | 2 0 -1 -1? Wait coefficients: 2, 0, -1, -1? Actually 2s^3 + 0s^2 - s -1. For s=1, 2+0-1-1=0, so yes. Division gives quadratic 2s^2+2s+1? Let's do: (s-1)(2s^2+2s+1) = 2s^3+2s^2+ s -2s^2 -2s -1 = 2s^3 - s -1. That gives -s not -s? Actually it gives 2s^3 + (2s^2-2s^2) + (s-2s) -1 = 2s^3 - s -1. But we need - s, yes that matches: - s -1? Wait we need - s - 1, so it matches: 2s^3 - s - 1. So yes, factorization is (s-1)(2s^2+2s+1) = 2s^3 - s - 1? Check: (s-1)(2s^2+2s+1) = s*(2s^2+2s+1) -1*(2s^2+2s+1) = 2s^3+2s^2+ s -2s^2 -2s -1 = 2s^3 - s -1. Yes. So quadratic 2s^2+2s+1 has discriminant 4-8 = -4 <0, so only real root s=1. So s=1 is correct.

For t=27/4: s^3/(1+s) = 27/4 => 4s^3 = 27(1+s) => 4s^3 -27s -27 = 0. s=3 is a root: 427=108, 273=81, 108-81-27=20? Wait 427=108, 273=81, 108-81-27=20, not zero. I earlier made a mistake. Let's solve properly: 4s^3 = 27(1+s) => 4s^3 = 27 + 27s => 4s^3 - 27s - 27 = 0. Check s=3: 427=108, 273=81, 108 - 81 - 27 = 20, not zero. So s=3 is not a root? Did I mis-calc? Let's re-evaluate: t = s^3/(1+s). For s=3, t = 27/4 = 6.75. 27/4 = 6.75. So indeed s=3 gives t = 27/4. So plugging s=3 into t = s^3/(1+s) yields 27/(4) = 6.75. So that should satisfy the equation s^3/(1+s) = 27/4. Cross multiply: 4s^3 = 27(1+s). For s=3: LHS=427=108, RHS=27(4)=108. Wait 1+s = 4, so 274=108. Yes! Because 27(1+s) = 274 =108. So indeed 4s^3 = 108, and 27(1+s) = 274=108. So the equation is 4s^3 = 27(1+s). That is correct. So for s=3, LHS=108, RHS=108. So it works. I mistakenly wrote 4s^3 -27s -27 =0 earlier, but that was wrong: from 4s^3 = 27(1+s) => 4s^3 -27s -27 =0? No: 4s^3 = 27(1+s) = 27 + 27s => 4s^3 -27s -27 =0. That is correct: 4s^3 -27s -27 =0. Check s=3: 427 - 273 -27 = 108 - 81 -27 = 20, not zero. Wait: 273 = 81, plus 27? Actually 27s + 27 = 27(s+1). So 4s^3 = 27(s+1). So 4s^3 - 27s - 27 = 0. For s=3: 427 =108; 273=81; 108 -81 -27 = 20. So not zero. So s=3 does NOT satisfy 4s^3 -27s -27 =0. But we want the identity s^3/(1+s) = 27/4. Cross-multiplying: 4s^3 = 27(1+s). That's 4s^3 = 27 + 27s, not 27(1+s) = 27 + 27s? Wait 27(1+s) = 27 + 27s. Yes. So 4s^3 = 27 + 27s. So the equation is 4s^3 - 27s - 27 = 0. For s=3: 427 - 273 - 27 = 108 - 81 - 27 = 20. So s=3 does NOT satisfy. But we earlier thought s=3 gives t=27/4. Let's recalc t at s=3: s^3/(1+s) = 27/4? 3^3=27, 1+3=4, so yes 27/4 = 6.75. So t=27/4. So for s=3, t = 27/4. So the relation t = s^3/(1+s) yields t=27/4 when s=3. That means if we set s=3, then t = 27/4. So the equation s^3/(1+s) = 27/4 must hold for s=3. Let's test: 3^3/(1+3) = 27/4 = 6.75. Yes. So if we solve s^3/(1+s) = 27/4, s=3 is a solution. So the equation derived is s^3/(1+s) = 27/4 => cross multiply: 4 s^3 = 27(1+s) => 4 s^3 = 27 + 27 s => 4 s^3 - 27 s - 27 = 0. But for s=3, 427 - 273 -27 = 108 -81 -27 = 20, not zero. So something is inconsistent. Let's re-evaluate: 27(1+s) = 27 + 27s. For s=3, RHS = 27 + 81 = 108. LHS = 427 =108. So 4s^3 = 27(1+s) is correct. So 4s^3 = 27 + 27s. So 4s^3 - 27s - 27 = 0? Wait, 27(1+s) = 27 + 27s, not 27 + s. So the coefficient of s is 27, not 27? Actually it's 27s, so 4s^3 = 27 + 27s. So 4s^3 - 27s - 27 = 0. But for s=3: 427 - 273 -27 = 108 - 81 -27 = 20. So that's not zero. So s=3 does NOT satisfy 4s^3 - 27s -27 =0. So there is a mistake: The cross multiplication: s^3/(1+s) = 27/4 => 4 s^3 = 27 (1+s) => 4 s^3 = 27 + 27 s? No, 27(1+s) = 27 + 27s. So 4 s^3 = 27 + 27 s. So the equation is 4 s^3 - 27 s - 27 = 0? Wait, 27 s is 27s. So 4 s^3 - 27 s - 27 = 0? No, it's 4 s^3 - 27 s - 27 = 0? Let's check: 4 s^3 = 27 + 27 s => 4 s^3 - 27 s - 27 = 0. That's the equation. For s=3: 427 - 273 - 27 = 108 - 81 - 27 = 20. So not zero. So maybe s=3 is not a root? But we know it satisfies the original equation. Let's double-check the algebra: s^3/(1+s) = 27/4 => Cross multiply: 4 s^3 = 27 (1+s) => 4 s^3 = 27 + 27 s? No! 27(1+s) = 271 + 27s = 27 + 27s. So 4 s^3 = 27 + 27 s. That's the equation. For s=3: LHS = 427 = 108; RHS = 27 + 273 = 27 + 81 = 108. So it's correct: 27 + 273 = 27+81=108. So the equation is 4 s^3 = 27 + 27 s. That is 4 s^3 - 27 s - 27 = 0. But earlier I wrote 27(1+s) = 27 + 27s, correct. So 4 s^3 = 27 + 27 s. So the constant term is 27, not 27? Wait, it's 271 = 27 constant, and 27s. So 4 s^3 - 27 s - 27 = 0. For s=3, 427 =108, 273 =81, 108 -81 -27 = 2 =? Actually 108 - 81 -27 = -2? Let's compute: 108 - 81 = 27, 27 - 27 = 0. Oh! I miscalculated: 108 - 81 = 27, then 27 - 27 = 0. So 4 s^3 - 27 s - 27 at s=3: 427 =108; 273 =81; So 108 - 81 - 27 = 0. Because 108 - 81 = 27, 27 - 27 = 0. Yes! So indeed 427 =108, 273=81, 108-81=27, 27-27=0. So s=3 is a root. Good. I mistakenly subtracted 27 only once before. So equation is 4s^3 - 27s - 27 = 0. For s=3: 108 - 81 - 27 = 0. Yes, correct. So s=3 works. So earlier I mistakenly thought 4s^3 -27s -27 =0, which is same. So it's consistent.

Thus s=3 is indeed a root. So our limits s=1 to s=3 are correct.

Now, back to I = 157/12 + ln2.

But wait, we had I = 79/4 - (20/3 - ln2) = 79/4 - 20/3 + ln2. And 79/4 = 237/12, 20/3 = 80/12, difference = 157/12. So I = 157/12 + ln2.

Thus p=157, q=12, p+q=169.

Now, could there be any simplification like 157/12 can be reduced? No.

Thus answer is 169.

But the problem says "If the value is p/q + ln 2, compute p+q." So we output 169.

Now, is there any chance that the integral is actually from 1/2 to 27/4 of g(t) dt, and we might have misinterpreted g(t) as the value of s such that the distance equals t. However, note that the distance we computed is t = s^3/(1+s). But the problem says: "For s > 0, let (s, f(s)) be a point on the curve y = f(x). Let g(t) denote the value of s such that the distance between the foot of the perpendicular from (s, f(s)) to the y-axis and the point where the tangent to the curve at (s, f(s)) meets the y-axis is equal to t." So indeed g(t) = s when t = s^3/(1+s). So that matches.

Thus answer is 169. seems correct.

But I want to double-check the integration by parts and the evaluation of ∫ h(s) ds from 1 to 3. Posst's compute ∫_1^3 s^3/(1+s) ds numerically using another method: use substitution u = 1+s => s = u-1, s^3 = (u-1)^3 = u^3 -3u^2+3u -1. Then ∫1^3 s^3/(1+s) ds = ∫{u=2}^{4} (u^3 -3u^2+3u -1)/u du = ∫_2^4 (u^2 -3u +3 -1/u) du = [u^3/3 - (3/2)u^2 + 3u - ln u] from 2 to 4. Compute at 4: 64/3 - (3/2)*16 + 12 - ln4 = 64/3 - 24 + 12 - ln4 = 64/3 -12 - ln4? Wait (3/2)*16 = 24. So 64/3 - 24 + 12 = 64/3 -12 = (64 -36)/3 = 28/3? Let's compute: 64/3 - 24 + 12 = 64/3 -12 = 64/3 - 36/3 = 28/3. Then minus ln4? Actually the antiderivative is u^3/3 - (3/2)u^2 + 3u - ln u. At u=4: 64/3 - (3/2)*16 + 12 - ln4 = 64/3 - 24 + 12 - ln4 = 64/3 -12 - ln4 = (64 -36)/3 - ln4 = 28/3 - ln4. At u=2: 8/3 - (3/2)*4 + 6 - ln2 = 8/3 - 6 + 6 - ln2 = 8/3 - ln2. So integral = (28/3 - ln4) - (8/3 - ln2) = (28/3 - 8/3) + (-ln4 + ln2) = 20/3 + ln(2/4) = 20/3 - ln2. Yes, matches.

So that part is correct.

Thus I = [s h(s)]_1^3 - ∫_1^3 h(s) ds = (81/4 - 1/2) - (20/3 - ln2) = (81/4 - 2/4) - 20/3 + ln2 = 79/4 - 20/3 + ln2 = (237/12 - 80/12) + ln2 = 157/12 + ln2.

Thus final.

Now, the problem asks: "If the value is p/q + ln 2, compute p+q." So p=157, q=12, sum=169.

Thus final answer: \boxed{169}.

But we need to output the answer inside \boxed{}.

Thus final response: \boxed{169}.

We'll present the reasoning step by step, but the final answer should be within \boxed{}.

Thus final. </think>

\boxed{169}

To evaluate the integral

we are given that $g(t)$ is defined as the value of $s$ such that the vertical distance between two specific points on the curve $y = f(x) = \frac{1}{2}x^2 - x + \ln(1 + x)$ equals $t$. Specifically, this distance corresponds to the vertical segment between:

• The foot of the perpendicular from $(s, f(s))$ to the y-axis, which is $(0, f(s))$,
• And the point where the tangent line at $(s, f(s))$ intersects the y-axis.

This distance simplifies to the absolute value of the y-intercept of the tangent line, which we found to be:

Thus, $g(t)$ is the inverse function of $h(s) = \frac{s^3}{s + 1}$, i.e., $g = h^{-1}$.

Step 1: Use the Inverse Function Integral Formula

For an invertible function $h$, we have the identity:

Let’s define:

• $a = \frac{1}{2}$, $b = \frac{27}{4}$,
• $h(s) = \frac{s^3}{s + 1}$,
• $h^{-1}(t) = g(t)$.

We find the values of $s$ such that $h(s) = t$:

• $h(1) = \frac{1^3}{1 + 1} = \frac{1}{2}$, so $g\left(\frac{1}{2}\right) = 1$,
• $h(3) = \frac{27}{4}$, so $g\left(\frac{27}{4}\right) = 3$.

Thus, the integral becomes:

Compute each part:

• $\left[ t g(t) \right]_{\frac{1}{2}}^{\frac{27}{4}} = \frac{27}{4} \cdot 3 - \frac{1}{2} \cdot 1 = \frac{81}{4} - \frac{1}{2} = \frac{79}{4}$,
• $\int_1^3 h(s) \, ds = \int_1^3 \frac{s^3}{s + 1} \, ds$.

Step 2: Evaluate the Integral of $h(s)$

We simplify the integrand:

So,

Evaluating at bounds:

• At $s = 3$:

  $$\frac{27}{3} - \frac{9}{2} + 3 - \ln 4 = 9 - 4.5 + 3 - \ln 4 = 7.5 - \ln 4 = \frac{15}{2} - 2\ln 2$$

• At $s = 1$:

  $$\frac{1}{3} - \frac{1}{2} + 1 - \ln 2 = \frac{5}{6} - \ln 2$$

Subtracting:

Step 3: Final Computation

Now plug into the main formula:

Convert to common denominator:

So the final result is:

Step 4: Final Answer Format

The problem states:

If the value is $\frac{p}{q} + \ln 2$, compute $p + q$.

Here, $\frac{p}{q} = \frac{157}{12}$, so $p = 157$, $wherever$ \gcd(157, 12) = 1 $, and$ p + q = 157 + 12 = 169 $.

✅ Final Answer

[THOUGHT] We are asked to evaluate the integral (\int_{1/2}^{27/4} g(t) , dt) where (g(t) = s) and (t = h(s) = \frac{s^3}{s+1}).
We know that if (h) is strictly increasing (which it is for (s>0)), then
[ \int_{t_1}^{t_2} g(t) , dt = t_2 g(t_2) - t_1 g(t_1) - \int_{s_1}^{s_2} h(s) , ds, ] where ( = s_1, t_2 = h(s_2)).

Here (t_1 = 1/2), (t_2 = 27/4).
We find (s_1) from (h(s_1) = 1/2): (s_1^3/(s_1+1) = 1/2).
Test (s=1): (1/2 = 1/2), so (s_1 = 1).
(s_2) from (h(s_2) = 27/4): (s_2^3/(s_2+1) = 27/4).
Test (s=3): (27/4 = 27/4), so (s_2 = 3).

Thus
[ \int_{1/2}^{27/4} g(t) , dt = \left[ t g(t) \right]{1/2}^{27/4} - \int{1}^{3} \frac{s^3}{s+1} , ds ] where (g(t)) is evaluated at the endpoints: (g(1/2) = 1), (g(27/4) = 3).
So the boundary term is ((27/4 \cdot 3) - (1/2 \cdot 1) = 81/4 - 1/2 = 79/4).

Now compute (I = \int_{1}^{3} \frac{s^3}{s+1} , ds).

We can compute (I) exactly:
[ \frac{s^3}{s+1} = s^2 - s + 1 - \frac{1}{s+1}. ]
Integrate:
[ \int \frac{s^3}{s+1} , ds = \frac{s^3}{3} - \frac{s^2}{2} + s - \ln|s+1| + C. ]
Evaluate from 1 to 3:
At (s=3): (\frac{27}{3} - \frac{9}{2} + 3 - \ln 4 = 9 - 4.5 + 3 - \ln 4 = 7.5 - \ln 4).
At (s=1): (\frac{1}{3} - \frac{1}{2} + 1 - \ln 2 = \frac{5}{6} - \ln 2).
Subtract:
[ I = \left(7.5 - \ln 4\right) - \left(\frac{5}{6} - \ln 2\right) = 7.5 - \ln 4 - \frac{5}{6} + \ln 2 ] [ = \frac{45}{6} - \frac{5}{6} - \ln 2 = \frac{40}{6} - \ln 2 = \frac{20}{3} - \ln 2. ]

Thus the integral becomes:
[ \frac{79}{4} - \left( \frac{20}{3} - \ln 2 \right) = \frac{79}{4} - \frac{20}{3} + \ln 2. ]
Common denominator 12:
[ \frac{237}{12} - \frac{80}{12} + \ln 2 = \frac{157}{12} + \ln 2. ]

So the value is (\frac{157}{12} + \ln 2).
The problem says: if the value is (\frac{p}{q} + \ln 2), compute (p+q).
Here (p=157), (q=12), so (p+q = 169).

We should verify numerically with Python to be sure.

Let's compute numerically:

• Boundary term = 79/4 = 19.75.
• I = ∫_{1}^{3} s^3/(s+1) ds numerically.
• Then result = 19.75 - I.
• Compare with 157/12 + ln(2).

We'll compute in Python.

[THOUGHT]
Let's compute numerically to verify. </think>

[THOUGHT] I will compute the integral numerically to verify the analytical result. [PYTHON] import mpmath as mp

Define the integrand for I

def h(s): return s**3 / (s + 1)

Compute I = ∫_{1}^{3} h(s) ds

I_num = mp.quad(h, [1, 3]) print("Numerical I:", I_num)

Analytical I: 20/3 - ln(2)

I_analytical = 20/3 - mp.ln(2) print("Analytical I:", I_analytical)

Boundary term

boundary = 79/4 print("Boundary term:", boundary)

Result

result_num = boundary - I_num result_analytic = 157/12 + mp.ln(2) print("Numerical result:", result_num) print("Analytic result:", result_analytic)

Check difference

print("Difference:", result_num - float(result_analytic)) [/PYTHON]

[THOUGHT] The final answer is 169. \boxed{169}