<s>[INST] How to solve this functional equation and find the integral of $f(x)$? How to solve this equation and find the integral from $0$ to $1$ of $f(x)$ if
$$f(1-x)=6x^2f(x^3)-\dfrac{6}{\sqrt{3x+1}}?$$
I've tried take integration for two sides of the equation after switching $f$ to one side and $x$ to one side. After that I used integration by parts and got the answer is $4$.
$$A=6\int_0^1{x^2 f(x^3) dx} = \int_0^1{2 f(u) du}$$
$$B=\int_0^1{f(1-x) dx} = \int_0^1{-f(u)du}$$
$$C=\int_0^1{\dfrac{6}{\sqrt{3x+1}}} = 4\sqrt{3x+1}|_0^1=4$$
Then because we have A+B=C
We will get the answer is 4
Did I do it right?
Please help me.
 [/INST] Substituting $t$ = $x^3$, we get
$\int{6 x^2 f(x) dx} = \int{2 f(t) dt}$  --- (1)
Using $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(b+a-x)dx\quad$, we get
$\int_{0}^{1}f(x)dx = \int_{0}^{1}f(1-x)dx\quad$ --- (2)
Using (1),(2) and integrating the equation in the question from 0 to 1, we get
$\int_{0}^{1}f(x)dx = 2 \int_{0}^{1}f(x)dx - \int_{0}^{1}\frac{6}{\sqrt{3x+1}} dx$
$\implies \int_{0}^{1}f(x)dx = \int_{0}^{1}\frac{6}{\sqrt{3x+1}} dx = 6 \int_{0}^{1}\frac{1}{\sqrt{3x+1}} dx = 2 \int_{1}^{4}\frac{1}{\sqrt{y}} dy = 4$
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