JEE Main 2024 — Definite Integration Question with Solution

$$\begin{array}{rl}& f(x)=\frac{x}{{\left(1+x^4\right)}^{1/4}}\\ & f\circ f(x)=\frac{f(x)}{{\left(1+f(x{)}^4\right)}^{1/4}}=\frac{\frac{x}{{\left(1+x^4\right)}^{1/4}}}{{\left(1+\frac{x^4}{1+x^4}\right)}^{1/4}}=\frac{x}{{\left(1+2x^4\right)}^{1/4}}\\ & f(f(f(f(x))))=\frac{x}{{\left(1+4x^4\right)}^{1/4}}\end{array}$$

18 \int_\limits0^{\sqrt{2 \sqrt{5}}} \frac{x^3}{\left(1+4 x^4\right)^{1 / 4}} d x

\begin{aligned} & \text { Let } 1+4 \mathrm{x}^4=\mathrm{t}^4 \\ & 16 \mathrm{x}^3 \mathrm{dx}=4 \mathrm{t}^3 \mathrm{dt} \\ & \frac{18}{4} \int_\limits1^3 \frac{\mathrm{t}^3 \mathrm{dt}}{\mathrm{t}} \\ & =\frac{9}{2}\left(\frac{\mathrm{t}^3}{3}\right)_1^3 \\ & =\frac{3}{2}[26]=39 \end{aligned}