<s>[INST] Find $\int_0^4(g\circ f\circ g)(x)\mathrm{d}x$ where $f(x)=\sqrt[3]{x+\sqrt{x^2+1/27}}+\sqrt[3]{x-\sqrt{x^2+1/27}}$, $g(x)=x^3+x+1$ 
Let $$f(x)=\sqrt[3]{x+\sqrt{x^2+\frac{1}{27}}}+\sqrt[3]{x-\sqrt{x^2+\frac{1}{27}}}$$
and
$$g(x)=x^3+x+1$$
then, find $$\int_0^4(g\circ f\circ g)(x) \mathrm dx$$

My attempt:
Let $\displaystyle h(x)=\sqrt{x^2+\frac{1}{27}}$
$$(g\circ f)(x)=2x+3((2x)(x^2-[h(x)]^2))^{1/3}+(x+h(x))^{1/3}+(x-h(x))^{1/3}+1$$
Is finding $(g\circ f \circ g)(x)$ in term of $x$ necessary? Because it is quite a work to do.
 [/INST] This is a nice problem.

$$f(x)=\underbrace{\sqrt[3]{x+\sqrt{x^2+\frac{1}{27}}}}_{a}+\underbrace{\sqrt[3]{x-\sqrt{x^2+\frac{1}{27}}}}_b$$
Now note that from the identity $(a+b)^3=a^3+b^3+3ab(a+b)$, and using $ab=-\dfrac{1}3$,
we have:
$$(f(x))^3=2x-f(x) \implies (f(x))^3+f(x)=2x $$
And from $g(x)=x^3+x+1$ , we have $$g(f(x))=(f(x))^3+f(x)+1=2x+1$$
Therefore $$g(f(g(x))=2g(x)+1=2x^3+2x+3$$
So all you have got is $$\int_0^4 (2x^3+2x+3)\mathrm{d}x$$
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