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.

Note that $f^{-1}(x) = \dfrac{x^3+x}{2}\\$
$\therefore g(x) = 2f^{-1}(x)+1\\$
$\therefore g(f(g(x))) = 2f^{-1}(f(g(x)))+1\\$
and I'm sure you know $f^{-1}(f(x)) = x \\$
$\therefore g(f(g(x))) = 2g(x)+1\\$
$$\therefore \int_0^4g(f(g(x)))dx = \int_0^4(2g(x)+1)dx = \int_0^4(2x^3+2x+3)dx\\$$
