<s>[INST] Let $f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ Then the value of $ \int^{3/4}_{1/4}f(f(x))\mathrm dx$ If  $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$
$\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 \mathrm dx$
could some help me with this, thanks
 [/INST] As @dxiv says in the comments, we can write $$f (x) = (x-\frac {1}{2})^3 +\frac {1}{4}x +\frac {3}{8} $$ Thus, substituting $u=x- \frac {1}{2}$ gives us $f (u)=u^3+\frac {1}{4}(u+\frac {1}{2}) +\frac {3}{8} = u^3+\frac {1}{4}u +\frac {1}{2} $. Thus, $$f (f (u)) =(u^3+\frac {1}{4}u +\frac {1}{2})^3 +\frac {1}{4}[u^3 +\frac {1}{4}u +\frac {1}{2}] +\frac {1}{2}$$ As $x $ goes from $\frac {1}{4} $ to $\frac {3}{4} $,. $u $ goes from $-\frac {1}{4} $ to $\frac {1}{4} $. 
Hope you can take it from here.
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