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Câu hỏi: cho hàm số $f(x)$ liên tục trên [0;1] và $f(x)+f(1-x)=\dfrac{{{x}^{2}}+2x+3}{x+1},\forall x\in \!\![\!\!0;1]$. Tính $\int\limits_{0}^{1}{f(x)}dx$
A. $\dfrac{3}{4}+2\ln 2$
B. $3+\ln 2$
C. $\dfrac{3}{4}+\ln 2$
D. $\dfrac{3}{2}+2\ln 2$
A. $\dfrac{3}{4}+2\ln 2$
B. $3+\ln 2$
C. $\dfrac{3}{4}+\ln 2$
D. $\dfrac{3}{2}+2\ln 2$
Lấy tích phân cận từ $0\to 1$ hai vế giả thiết, ta được $\int\limits_{0}^{1}{f\left( x \right)dx}+\int\limits_{0}^{1}{f\left( 1-x \right)dx}=\int\limits_{0}^{1}{\dfrac{{{x}^{2}}+2x+3}{x+1}dx}$
Lại có: $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( 1-x \right)dx}$
Do đó: $\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{1}{\left( x+1+\dfrac{2}{x+1} \right)dx=}\dfrac{1}{2}\left( \dfrac{{{x}^{2}}}{2}+x+2\ln \left| x+1 \right| \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{3}{4}+\ln 2$.
Lại có: $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( 1-x \right)dx}$
Do đó: $\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{1}{\left( x+1+\dfrac{2}{x+1} \right)dx=}\dfrac{1}{2}\left( \dfrac{{{x}^{2}}}{2}+x+2\ln \left| x+1 \right| \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{3}{4}+\ln 2$.
Đáp án C.