Giải thích các bước giải:
Đặt $2x-1=t\to x=\dfrac{t+1}{2}$
$\to f'(t)=(\dfrac{t+1}{2}+1)(\dfrac{t+1}{2}-2)^2(\dfrac{t+1}{2}-3)$
$\to f'(t)=\dfrac{t+3}{2}.(\dfrac{t-1}{2})^2.\dfrac{t-5}{2}$
$\to f'(x)=\dfrac{x+3}{2}.(\dfrac{x-1}{2})^2.\dfrac{x-5}{2}$
$\to f'(x^2+1)=2x.f'(x^2+1)=2x.\dfrac{x^2+1+3}{2}.(\dfrac{x^2+1-1}{2})^2.\dfrac{x^2+1-5}{2}$
$\to f'(x^2+1)=2x.\dfrac{x^2+4}{2}.(\dfrac{x^2}{2})^2.\dfrac{x^2-4}{2}$
$\to f'(x^2+1)=2x.\dfrac{x^2+4}{2}.(\dfrac{x^2}{2})^2.\dfrac{(x-2)(x+2)}{2}$
$\to f(x^2+1)$ đồng biến trên : $[-2,0]\cup[2,+\infty)$
$\to C$
