Let $$f: R \rightarrow R$$ be a function defined $$f(x) = \frac{x}{(1+x^4)^{1/4}}$$ and $$g(x) = f(f(f(f(x))))$$ then $$18\int_0^{\sqrt{2\sqrt{5}}} x^2 g(x) \, dx$$

We start with the function $$f:\mathbb{R}\rightarrow\mathbb{R}$$ defined by $$f(x)=\dfrac{x}{(1+x^4)^{1/4}}\,.$$

Step 1 : An equation relating $$f(x)$$ and $$x$$.
Let $$y=f(x)\,.$$ Then

$$y=\dfrac{x}{(1+x^4)^{1/4}} \;\;\Longrightarrow\;\; y^4=\dfrac{x^4}{1+x^4}\;.$$ $$-(1)$$

Re-arranging $$-(1)$$ to express $$x^4$$ in terms of $$y^4$$:

$$y^4(1+x^4)=x^4 \;\;\Longrightarrow\;\; y^4=x^4(1-y^4) \;\;\Longrightarrow\;\; x^4=\dfrac{y^4}{1-y^4}\;.$$ $$-(2)$$

Equation $$-(2)$$ shows that applying $$f$$ replaces the quantity $$t=x^4$$ by the new value $$\dfrac{t}{1+t}\,. $$

Step 2 : Iterating the transformation.
Define $$T(t)=\dfrac{t}{1+t}\;.$$
If $$t_0=x^4$$, then after one application of $$f$$ we have $$t_1=T(t_0)$$, after two applications $$t_2=T^2(t_0)$$, and so on.

Compute the first four iterates:

$$T(t)=\dfrac{t}{1+t}\,,$$

$$T^2(t)=T\!\bigl(T(t)\bigr)=\dfrac{t}{1+2t}\,,$$

$$T^3(t)=T\!\bigl(T^2(t)\bigr)=\dfrac{t}{1+3t}\,,$$

$$T^4(t)=T\!\bigl(T^3(t)\bigr)=\dfrac{t}{1+4t}\,. $$

Step 3 : Simplifying $$g(x)=f^{(4)}(x).$$
Setting $$t_0=x^4$$, after four applications we get

$$g(x)^4 = T^4(x^4)=\dfrac{x^4}{1+4x^4}\;.$$

Hence

$$g(x)=\dfrac{x}{(1+4x^4)^{1/4}}\,. $$ $$-(3)$$

Step 4 : Setting up the required integral.
The integrand becomes

$$x^2\,g(x)=x^2\cdot\dfrac{x}{(1+4x^4)^{1/4}}=\dfrac{x^3}{(1+4x^4)^{1/4}}\;.$$

Therefore

$$I=\displaystyle\int_{0}^{\sqrt{2\sqrt{5}}}\dfrac{x^3}{(1+4x^4)^{1/4}}\,dx\;.$$ $$-(4)$$

Step 5 : Substitution to evaluate $$I$$.
Put $$u=1+4x^4\;,$$ so that $$\dfrac{du}{dx}=16x^3\;\Longrightarrow\;x^3\,dx=\dfrac{du}{16}\,.$$

When $$x=0$$, $$u=1$$;  when $$x=\sqrt{2\sqrt{5}}$$,

$$x^4=(\sqrt{2\sqrt{5}})^4=(2\sqrt{5})^2=4\cdot5=20 \;\Longrightarrow\; u=1+4\cdot20=81\;.$$

Substituting in $$-(4)$$:

$$I=\dfrac{1}{16}\displaystyle\int_{1}^{81}u^{-1/4}\,du\;.$$

Use the power-rule integral $$\int u^{n}\,du=\dfrac{u^{n+1}}{n+1}+C\;.$$
With $$n=-\dfrac14$$ we get

$$\int u^{-1/4}\,du=\dfrac{u^{3/4}}{3/4}=\dfrac{4}{3}u^{3/4}\;.$$

Hence

$$I=\dfrac{1}{16}\cdot\dfrac{4}{3}\Bigl[u^{3/4}\Bigr]_{1}^{81} =\dfrac{1}{12}\Bigl(81^{3/4}-1^{3/4}\Bigr)\;.$$

Because $$81=3^4,\;81^{1/4}=3,\;81^{3/4}=3^3=27$$, we have

$$I=\dfrac{1}{12}(27-1)=\dfrac{26}{12}=\dfrac{13}{6}\;.$$

Step 6 : Final multiplication.
The question asks for $$18I$$:

$$18I=18\cdot\dfrac{13}{6}=3\cdot13=39\;.$$

Answer : 39   (Option D)