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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa mãn $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( \tan x \right)dx}=3$ và $\int\limits_{0}^{1}{\dfrac{{{x}^{2}}f\left( x \right)}{{{x}^{2}}+1}dx}=1$. Tính $I=\int\limits_{0}^{1}{f\left( x \right)dx}$
A. $I=3$.
B. $I=2$.
C. $I=4$.
D. $I=6$.
A. $I=3$.
B. $I=2$.
C. $I=4$.
D. $I=6$.
Xét ${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( \tan x \right)dx}=3$
Đặt $t=\tan x\Rightarrow dt=\dfrac{1}{{{\cos }^{2}}x}dx$
${{I}_{1}}=\int\limits_{0}^{1}{f\left( t \right).\dfrac{1}{1+{{t}^{2}}}dt}=\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{1+{{x}^{2}}}dx}=3$
Ta có:
$\begin{aligned}
& \int\limits_{0}^{1}{\dfrac{{{x}^{2}}f\left( x \right)}{{{x}^{2}}+1}dx}=1\Leftrightarrow \int\limits_{0}^{1}{\left( 1-\dfrac{1}{{{x}^{2}}+1} \right)f\left( x \right)}=1\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)dx}-\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{{{x}^{2}}+1}dx}=1 \\
& \Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)dx}-3=1\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=4 \\
\end{aligned}$
Đặt $t=\tan x\Rightarrow dt=\dfrac{1}{{{\cos }^{2}}x}dx$
${{I}_{1}}=\int\limits_{0}^{1}{f\left( t \right).\dfrac{1}{1+{{t}^{2}}}dt}=\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{1+{{x}^{2}}}dx}=3$
Ta có:
$\begin{aligned}
& \int\limits_{0}^{1}{\dfrac{{{x}^{2}}f\left( x \right)}{{{x}^{2}}+1}dx}=1\Leftrightarrow \int\limits_{0}^{1}{\left( 1-\dfrac{1}{{{x}^{2}}+1} \right)f\left( x \right)}=1\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)dx}-\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{{{x}^{2}}+1}dx}=1 \\
& \Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)dx}-3=1\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=4 \\
\end{aligned}$
Đáp án C.