Cho hàm số \(f(x)\) liên tục trên \(\mathbb{R}\) và \[\int\limits_0^{\frac{\pi }{4}} {f\left( {\tan x} \right)dx} = \int\limits_0^1 {\frac{{{x^2}f\left( x \right)}}{{{x^2} + 1}}} \,{\rm{d}}x = 2.\] Tính \(I = \int\limits_0^1 {f\left( x \right)} \,{\rm{d}}x.\)

Đặt \(u = \tan x \Rightarrow du = \frac{1}{{{{\cos }^2}x}}dx = \left( {1 + {{\tan }^2}x} \right)dx \Rightarrow \frac{{{u^2} + 1}} = dx.\)

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow u = 0\\x = \frac{\pi }{4} \Rightarrow u = 1\end{array} \right.\).

Ta có: \(\int\limits_0^{\frac{\pi }{4}} {f\left( {\tan x} \right)dx}  = \int\limits_0^1 {\frac{{f\left( u \right)}}{{{u^2} + 1}}} \,{\rm{d}}u = \int\limits_0^1 {\frac{{f\left( x \right)}}{{{x^2} + 1}}} \,{\rm{d}}x \Rightarrow \int\limits_0^1 {\frac{{f\left( x \right)}}{{{x^2} + 1}}} \,{\rm{d}}x = 2\).

Do đó \[I = \int\limits_0^1 {f\left( x \right)} \,{\rm{d}}x = \int\limits_0^1 {\frac{{\left( {{x^2} + 1} \right)f\left( x \right)}}{{{x^2} + 1}}} \,{\rm{d}}x = \int\limits_0^1 {\frac{{{x^2}f\left( x \right)}}{{{x^2} + 1}}\,} dx + \int\limits_0^1 {\frac{{f\left( x \right)}}{{{x^2} + 1}}} \,\,dx = 2 + 2 = 4\].

Chọn C.