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Câu hỏi: Cho tích phân $I=\int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{\dfrac{x\ln \left( \dfrac{1+x}{1-x} \right)}{{{e}^{x}}+1}dx}=a\ln b+c$ thì giá trị của $a-b+c$ là:
A. $a-b+c=-\dfrac{23}{8}.$
B. $a-b+c=-\dfrac{17}{8}.$
C. $a-b+c=\dfrac{31}{8}.$
D. $a-b+c=\dfrac{23}{8}.$
A. $a-b+c=-\dfrac{23}{8}.$
B. $a-b+c=-\dfrac{17}{8}.$
C. $a-b+c=\dfrac{31}{8}.$
D. $a-b+c=\dfrac{23}{8}.$
Vì hàm số $f\left( x \right)=x\ln \left( \dfrac{1+x}{1-x} \right)$ là hàm số chẵn và liên tục trên $\left[ -\dfrac{1}{2};\dfrac{1}{2} \right]$ nên ta có:
$I=\int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{\dfrac{x\ln \left( \dfrac{1+x}{1-x} \right)}{{{e}^{x}}+1}dx}=\int\limits_{0}^{\dfrac{1}{2}}{x\ln \left( \dfrac{1+x}{1-x} \right)dx}$.
Đặt $\left\{ \begin{aligned}
& u=\ln \left( \dfrac{1+x}{1-x} \right) \\
& dv=xdx \\
\end{aligned} \right. $, ta có: $ \left\{ \begin{aligned}
& du=\dfrac{-2}{{{x}^{2}}-1} \\
& v=\dfrac{1}{2}\left( {{x}^{2}}-1 \right) \\
\end{aligned} \right.$.
$I=\dfrac{1}{2}\left( {{x}^{2}}-1 \right)\ln \left( \dfrac{1+x}{1-x} \right)\left| \begin{aligned}
& ^{\dfrac{1}{2}} \\
& _{0} \\
\end{aligned} \right.+\int\limits_{0}^{\dfrac{1}{2}}{dx}=-\dfrac{3}{8}\ln 3+\dfrac{1}{2}\Rightarrow a=-\dfrac{3}{8};b=3;c=\dfrac{1}{2}$.
Vậy $a-b+c=-\dfrac{3}{8}-3+\dfrac{1}{2}=-\dfrac{23}{8}.$
$I=\int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{\dfrac{x\ln \left( \dfrac{1+x}{1-x} \right)}{{{e}^{x}}+1}dx}=\int\limits_{0}^{\dfrac{1}{2}}{x\ln \left( \dfrac{1+x}{1-x} \right)dx}$.
Đặt $\left\{ \begin{aligned}
& u=\ln \left( \dfrac{1+x}{1-x} \right) \\
& dv=xdx \\
\end{aligned} \right. $, ta có: $ \left\{ \begin{aligned}
& du=\dfrac{-2}{{{x}^{2}}-1} \\
& v=\dfrac{1}{2}\left( {{x}^{2}}-1 \right) \\
\end{aligned} \right.$.
$I=\dfrac{1}{2}\left( {{x}^{2}}-1 \right)\ln \left( \dfrac{1+x}{1-x} \right)\left| \begin{aligned}
& ^{\dfrac{1}{2}} \\
& _{0} \\
\end{aligned} \right.+\int\limits_{0}^{\dfrac{1}{2}}{dx}=-\dfrac{3}{8}\ln 3+\dfrac{1}{2}\Rightarrow a=-\dfrac{3}{8};b=3;c=\dfrac{1}{2}$.
Vậy $a-b+c=-\dfrac{3}{8}-3+\dfrac{1}{2}=-\dfrac{23}{8}.$
Đáp án A.