Google
Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ và $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( \tan x \right)\text{d}x}=\int\limits_{0}^{1}{\dfrac{{{x}^{2}}f\left( x \right)}{{{x}^{2}}+1}}\text{d}x=2.$
Tính $I=\int\limits_{0}^{1}{f\left( x \right)}\text{d}x$
A. $I=-4$.
B. $I=2$.
C. $I=4$.
D. $I=6$.
Tính $I=\int\limits_{0}^{1}{f\left( x \right)}\text{d}x$
A. $I=-4$.
B. $I=2$.
C. $I=4$.
D. $I=6$.
Đặt $u=\tan x\Rightarrow \text{d}u=\dfrac{1}{{{\cos }^{2}}x}\text{d}x=\left( 1+{{\tan }^{2}}x \right)dx\Rightarrow \dfrac{\text{d}u}{{{u}^{2}}+1}=\text{d}x$
Đổi cận:
Ta có: $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( \tan x \right)\text{d}x}=\int\limits_{0}^{1}{\dfrac{f\left( u \right)}{{{u}^{2}}+1}}\text{d}u=\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{{{x}^{2}}+1}}\text{d}x\Rightarrow \int\limits_{0}^{1}{\dfrac{f\left( x \right)}{{{x}^{2}}+1}}\text{d}x=2$.
$I=\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=\int\limits_{0}^{1}{\dfrac{\left( {{x}^{2}}+1 \right)f\left( x \right)}{{{x}^{2}}+1}}\text{d}x=\int\limits_{0}^{1}{\dfrac{{{x}^{2}}f\left( x \right)}{{{x}^{2}}+1}}\text{d}x+\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{{{x}^{2}}+1}}\text{d}x=2+2=4.$
Đổi cận:
$I=\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=\int\limits_{0}^{1}{\dfrac{\left( {{x}^{2}}+1 \right)f\left( x \right)}{{{x}^{2}}+1}}\text{d}x=\int\limits_{0}^{1}{\dfrac{{{x}^{2}}f\left( x \right)}{{{x}^{2}}+1}}\text{d}x+\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{{{x}^{2}}+1}}\text{d}x=2+2=4.$
Đáp án C.