Let f: R → R be a function defined f(x) = x (1 + x ^ 4 ) ^ 1 / 4...

Given,

$f\left(x\right) = \dfrac{x}{{\left(1 + {x}^{4}\right)}^{1 / 4}}$

Now, finding

$fof\left(x\right) = \dfrac{f\left(x\right)}{{\left(1 + {f}^{4}\left(x\right)\right)}^{1 / 4}} = \dfrac{\dfrac{x}{{\left(1 + {x}^{4}\right)}^{1 / 4}}}{{\left(1 + \dfrac{{x}^{4}}{1 + {x}^{4}}\right)}^{1 / 4}} = \dfrac{x}{{\left(1 + 2{x}^{4}\right)}^{1 / 4}}$

And $f\left(f\left(f\left(x\right)\right)\right) = \dfrac{f\left(x\right)}{{\left(1 + 2{f}^{4}\left(x\right)\right)}^{1 / 4}} = \dfrac{\dfrac{x}{{\left(1 + {x}^{4}\right)}^{1 / 4}}}{{\left(1 + \dfrac{2{x}^{4}}{1 + {x}^{4}}\right)}^{1 / 4}} = \dfrac{x}{{\left(1 + 3{x}^{4}\right)}^{1 / 4}}$

$⇒ f\left(f\left(f\left(f\left(x\right)\right)\right)\right) = \dfrac{x}{{\left(1 + 4{x}^{4}\right)}^{1 / 4}} = g\left(x\right)$

Now finding,

$18{∫}_{0}^{\sqrt{2\sqrt{5}}}{x}^{2}g\left(x\right)dx$

$= 18{∫}_{0}^{\sqrt{2\sqrt{5}}}\dfrac{{x}^{3}}{{\left(1 + 4{x}^{4}\right)}^{1 / 4}}dx$

Let $1 + 4{x}^{4} = {t}^{4}$

$⇒ 16{x}^{3}dx = 4{t}^{3}dt$

So, the integral becomes

$\dfrac{18}{4}{∫}_{1}^{3}\dfrac{{t}^{3}dt}{t}$

$= \dfrac{9}{2}{∫}_{1}^{3}{t}^{2}dt$

$= \dfrac{9}{2}{\left(\dfrac{{t}^{3}}{3}\right)}_{1}^{3}$

$= \dfrac{3}{2} \cdot 26 = 39$