A polynomial Calculus function

$\begin{aligned} I & = \int_\frac 14^\frac 34 f\left(f(x)\right) \ dx & \small \color{#3D99F6} \text{Using the identity: }\int_a^b f(x) \ dx = \int_a^b f(a+ b-x) \ dx \\ & = \frac 12 \int_\frac 14^\frac 34 f\left(f(x)\right) + {\color{#3D99F6}f\left(f(1-x)\right)} \ dx & \small \color{#3D99F6} \text{See note: }f(1-x) = 1-f(x) \\ & = \frac 12 \int_\frac 14^\frac 34 f\left(f(x)\right) + {\color{#3D99F6}f\left(1-f(x)\right)} \ dx & \small \color{#3D99F6} \text{Again }f(1-f(x)) = 1-f\left(f(x)\right) \\ & = \frac 12 \int_\frac 14^\frac 34 f\left(f(x)\right) + {\color{#3D99F6}1 - f\left(f(x)\right)} \ dx \\ & = \frac 12 \int_\frac 14^\frac 34 \ dx = \frac 14 \\ \implies \frac 1I & = \boxed{4} \end{aligned}$

Note:

$\small \begin{aligned} f(x) & = x^3-\frac 32x^2 + x + \frac 14 \\ \implies f(1-x) & = \frac 34 - x + \frac 32x^2 - x^3 = 1 - \left(x^3-\frac 32x^2 + x + \frac 14\right) = 1 - f(x) \end{aligned}$