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

The base of the solid is bounded by:

• The curve $y = \sqrt{x + x \ln x}$
• The x-axis
• The vertical lines $x = 1$ and $x = 2$

Cross-sections perpendicular to the x-axis are equilateral triangles.

For an equilateral triangle with side length $s$, the area is $\frac{\sqrt{3}}{4}s^2$.

At a given $x$, the side length of the equilateral triangle is the distance from the x-axis to the curve, which is $y = \sqrt{x + x \ln x}$.

So the side length $s(x) = \sqrt{x + x \ln x} = \sqrt{x(1 + \ln x)}$.

The area of the cross-section at position $x$ is: $A(x) = \frac{\sqrt{3}}{4} s(x)^2 = \frac{\sqrt{3}}{4} \cdot x(1 + \ln x)$

The volume is: $V = \int_1^2 A(x) \, dx = \frac{\sqrt{3}}{4} \int_1^2 x(1 + \ln x) \, dx$

Let me compute this integral. [/THOUGHT] [PYTHON] from sympy import symbols, sqrt, ln, integrate, simplify, Rational

x = symbols('x')

The side length squared

s_squared = x * (1 + ln(x))

Area of equilateral triangle with side s is (sqrt(3)/4) * s^2

So A(x) = (sqrt(3)/4) * x * (1 + ln(x))

Volume = integral from 1 to 2 of A(x) dx

V = (sqrt(3)/4) * integral from 1 to 2 of x*(1 + ln(x)) dx

Let's compute the integral of x*(1 + ln(x)) from 1 to 2

integrand = x * (1 + ln(x)) integral_result = integrate(integrand, (x, 1, 2)) print("Integral of x*(1 + ln(x)) from 1 to 2:", integral_result) print("Simplified:", simplify(integral_result))

Volume = (sqrt(3)/4) * integral_result

volume = (sqrt(3)/4) * integral_result volume_simplified = simplify(volume) print("Volume:", volume_simplified)

Let's expand the integral manually to verify

integral of x dx = x^2/2

integral of x*ln(x) dx: use integration by parts

Let u = ln(x), dv = x dx

du = 1/x dx, v = x^2/2

integral = (x^2/2)ln(x) - integral of (x^2/2)(1/x) dx

= (x^2/2)*ln(x) - integral of x/2 dx

= (x^2/2)*ln(x) - x^2/4

So integral of x*(1 + ln(x)) = x^2/2 + (x^2/2)*ln(x) - x^2/4

= x^2/4 + (x^2/2)*ln(x)

= (x^2/4)(1 + 2ln(x))

Evaluating from 1 to 2:

At x=2: (4/4)(1 + 2ln(2)) = 1 + 2*ln(2)

At x=1: (1/4)(1 + 2ln(1)) = (1/4)*(1 + 0) = 1/4

Result: (1 + 2ln(2)) - 1/4 = 3/4 + 2ln(2)

print("\nManual calculation:") at_2 = (4/4)(1 + 2ln(2)) at_1 = Rational(1,4)(1 + 2ln(1)) manual_integral = at_2 - at_1 print("At x=2:", at_2) print("At x=1:", at_1) print("Integral:", simplify(manual_integral))

Volume = (sqrt(3)/4) * (3/4 + 2*ln(2))

= sqrt(3) * (3/4 + 2*ln(2)) / 4

= sqrt(3) * (3 + 8*ln(2)) / 16

volume_final = sqrt(3) * (Rational(3,4) + 2ln(2)) / 4 print("\nVolume = sqrt(3) * (3/4 + 2ln(2)) / 4") print(" = sqrt(3) * (3 + 8*ln(2)) / 16") print("Simplified volume:", simplify(volume_final))

So the form is sqrt(3)(p + qln(2))/r

where p = 3, q = 8, r = 16

p, q, r = 3, 8, 16 print(f"\np = {p}, q = {q}, r = {r}") print(f"p + q + r = {p + q + r}") [/PYTHON]

[THOUGHT] The calculation confirms that the volume is $\frac{\sqrt{3}(3 + 8\ln 2)}{16}$.

Comparing with the given form $\frac{\sqrt{3}(p + q\ln 2)}{r}$, we have:

• $p = 3$
• $q = 8$
• $r = 16$

Therefore, $p + q + r = 3 + 8 + 16 = 27$. [/THOUGHT]

\boxed{27}