If $f(x) = \frac{x^3}{3} -\frac{x^2}{2} + x + \frac{1}{12}$, then $\int_{\frac{1}{7}}^{\frac{6}{7}}f(f(x))\,dx =\,$? This is a question from a practice workbook for a college entrance exam.

Let
$$f(x) = \frac{x^3}{3} -\frac{x^2}{2} + x + \frac{1}{12}.$$
Find
$$\int_{\frac{1}{7}}^{\frac{6}{7}}f(f(x))\,dx.$$

While I know that computing $f(f(x))$ is an option, it is very time consuming and wouldn't be practical considering the time limit of the exam. I believe there must be a more elegant solution.
Looking at the limits, I tried to find useful things about $f(\frac{1}{7}+\frac{6}{7}-x)$
The relation I obtained was that $f(x) + f(1-x) = 12/12 = 1$. I don't know how to use this for the direct integral of $f(f(x)).$

In a comment after @Math Lover's elagant answer, I told that nobody tried the change of variable $f(x)=y$. So, I tried for the fun of it
$$I=\int f[f(x)]\,dx=\int \Big[ \frac 1{12}+f(x)-\frac 12 f^2(x)+\frac 13 f^3(x)\Big]\,dx$$
Let $f(x)=y$. Solving the cubic with the hyperbolic method for only one real root
$$x=\frac{1}{2}-\sqrt{3} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{2-4
   y}{\sqrt{3}}\right)\right)$$
$$dx=\frac{4 \cosh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{2-4
   y}{\sqrt{3}}\right)\right)}{\sqrt{48 (y-1) y+21}}\,dy$$ So,
$$20480\,I=-10240 \sqrt{3} \sinh (t)+2700 \cosh (2 t)+1350 \cosh (4 t)+15 \cosh (8 t)+12 \cosh   (10 t)$$ where $$t=\frac{1}{3} \sinh ^{-1}\left(\frac{2-4 y}{\sqrt{3}}\right)$$ Now, for $y$, the upper and lower bounds are $\frac{3223}{4116}$ and $\frac{893}{4116}$ and then for $t$, they are $$\mp \frac{1}{3} \sinh ^{-1}\left(\frac{1165}{1029 \sqrt{3}}\right)$$ Because of the symmetry in $t$, all cosines disappear and the result for the definite integral is just
$$\int_{\frac{1}{7}}^{\frac{6}{7}}f[f(x)]\,dx =\sqrt{3} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{1165}{1029 \sqrt{3}}\right)\right)$$ whic, numerically is $0.3571428571428571428571429$; its reciprocal is $2.8=\frac {14}5$ so the value of $\frac 5{14}$.
For those who are curious, this took me close to one and half hour.
