Cho hàm số f(x) liên tục trên R thỏa mãn

A

- Hướng dẫn giải

Ta có \({I_1} = \int\limits_0^{\frac{\pi }{4}} {f\left( {\tan x} \right)} \,{\rm{d}}x = 4\).

Đặt \(t = \tan x \Rightarrow {\rm{d}}t = \frac{{{\rm{d}}x}}{{{{\cos }^2}x}} \Rightarrow {\rm{d}}t = \left( {1 + {{\tan }^2}x} \right){\rm{d}}x = \left( {1 + {t^2}} \right){\rm{d}}x \Rightarrow \frac{{{\rm{d}}t}}{{1 + {t^2}}} = {\rm{d}}x\).

\( \Rightarrow {\rm{d}}t = \left( {1 + {{\tan }^2}x} \right){\rm{d}}x = \left( {1 + {t^2}} \right){\rm{d}}x \Rightarrow \frac{{{\rm{d}}t}}{{1 + {t^2}}} = {\rm{d}}x\).

\( \Rightarrow {I_1} = \int\limits_0^1 {\frac{{f(t)}}{{{t^2} + 1}}{\rm{d}}t} = \int\limits_0^1 {\frac{{f(x)}}{{{x^2} + 1}}{\rm{d}}x} = 4\).

\({I_2} = \int\limits_0^1 {\frac{{{x^2}f\left( x \right)}}{{{x^2} + 1}}\,} {\rm{d}}x = \int\limits_0^1 {f(x)} \,{\rm{d}}x - \int\limits_0^1 {\frac{{f\left( x \right)}}{{{x^2} + 1}}} \,{\rm{d}}x = \int\limits_0^1 {f\left( x \right)} \,{\rm{d}}x - 4 = 2 \Rightarrow \int\limits_0^1 {f\left( x \right)} \,{\rm{d}}x = 6\)