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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{aligned}
& 3{{x}^{2}}\ln \left( x+1 \right)\text{ khi }x\ge 0 \\
& 2x\sqrt{{{x}^{2}}+3}+1\text{ khi }x<0 \\
\end{aligned} \right. $. Biết $ \int\limits_{\dfrac{1}{e}}^{e}{\dfrac{f\left( \ln x \right)}{x}dx}=a\sqrt{3}+b\ln 2+c $ với $ a,b,c\in \mathbb{Q} $. Giá trị của $ a+b+6c$ bằng
A. $35$.
B. $-14$.
C. $-27$.
D. $18$.
& 3{{x}^{2}}\ln \left( x+1 \right)\text{ khi }x\ge 0 \\
& 2x\sqrt{{{x}^{2}}+3}+1\text{ khi }x<0 \\
\end{aligned} \right. $. Biết $ \int\limits_{\dfrac{1}{e}}^{e}{\dfrac{f\left( \ln x \right)}{x}dx}=a\sqrt{3}+b\ln 2+c $ với $ a,b,c\in \mathbb{Q} $. Giá trị của $ a+b+6c$ bằng
A. $35$.
B. $-14$.
C. $-27$.
D. $18$.
Đặt $t=\ln x$, $dt=\dfrac{1}{x}dx$ và $x=\dfrac{1}{e}$ $\Rightarrow t=-1$ ; $x=e$ $\Rightarrow t=1$.
$\int\limits_{\dfrac{1}{e}}^{e}{\dfrac{f\left( \ln x \right)}{x}dx}=\int\limits_{-1}^{1}{f\left( t \right)dt}=\int\limits_{-1}^{0}{\left( 2t\sqrt{{{t}^{2}}+3}+1 \right)dt}+\int\limits_{0}^{1}{3{{t}^{2}}\ln \left( t+1 \right)dt}$.
Tính $I=\int\limits_{-1}^{0}{\left( 2t\sqrt{{{t}^{2}}+3}+1 \right)dt}=\int\limits_{-1}^{0}{2t\sqrt{{{t}^{2}}+3}dt}+\int\limits_{-1}^{0}{dt}=\int\limits_{-1}^{0}{2t\sqrt{{{t}^{2}}+3}dt}+1$.
Đặt $u=\sqrt{{{t}^{2}}+3}$ $\Rightarrow udu=tdt$ ; $t=-1\Rightarrow u=2;t=0\Rightarrow u=\sqrt{3}$.
$I=\int\limits_{2}^{\sqrt{3}}{2{{u}^{2}}du}+1=\left. 2\dfrac{{{u}^{3}}}{3} \right|_{2}^{\sqrt{3}}+1=2\sqrt{3}-\dfrac{13}{3}$.
Tính $J=\int\limits_{0}^{1}{3{{t}^{2}}\ln \left( t+1 \right)dt}$
$\left\{ \begin{aligned}
& u=\ln \left( t+1 \right) \\
& dv={{t}^{2}}dt \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{t+1}dt \\
& v=\dfrac{{{t}^{3}}+1}{3} \\
\end{aligned} \right.$
$J=\left. \left( {{t}^{3}}+1 \right)\ln \left( t+1 \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\left( {{t}^{2}}-t+1 \right)dt}$ $=\left. \left( {{t}^{3}}+1 \right)\ln \left( t+1 \right) \right|_{0}^{1}-\left. \left( \dfrac{{{t}^{3}}}{3}-\dfrac{{{t}^{2}}}{2}+t \right) \right|_{0}^{1}=2\ln 2-\dfrac{5}{6}$.
Vậy $\int\limits_{\dfrac{1}{e}}^{e}{\dfrac{f\left( \ln x \right)}{x}dx}=2\sqrt{3}+2\ln 2-\dfrac{31}{6}$.
$\Rightarrow $ $a+b+6c=2+2-31=-27$.
$\int\limits_{\dfrac{1}{e}}^{e}{\dfrac{f\left( \ln x \right)}{x}dx}=\int\limits_{-1}^{1}{f\left( t \right)dt}=\int\limits_{-1}^{0}{\left( 2t\sqrt{{{t}^{2}}+3}+1 \right)dt}+\int\limits_{0}^{1}{3{{t}^{2}}\ln \left( t+1 \right)dt}$.
Tính $I=\int\limits_{-1}^{0}{\left( 2t\sqrt{{{t}^{2}}+3}+1 \right)dt}=\int\limits_{-1}^{0}{2t\sqrt{{{t}^{2}}+3}dt}+\int\limits_{-1}^{0}{dt}=\int\limits_{-1}^{0}{2t\sqrt{{{t}^{2}}+3}dt}+1$.
Đặt $u=\sqrt{{{t}^{2}}+3}$ $\Rightarrow udu=tdt$ ; $t=-1\Rightarrow u=2;t=0\Rightarrow u=\sqrt{3}$.
$I=\int\limits_{2}^{\sqrt{3}}{2{{u}^{2}}du}+1=\left. 2\dfrac{{{u}^{3}}}{3} \right|_{2}^{\sqrt{3}}+1=2\sqrt{3}-\dfrac{13}{3}$.
Tính $J=\int\limits_{0}^{1}{3{{t}^{2}}\ln \left( t+1 \right)dt}$
$\left\{ \begin{aligned}
& u=\ln \left( t+1 \right) \\
& dv={{t}^{2}}dt \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{1}{t+1}dt \\
& v=\dfrac{{{t}^{3}}+1}{3} \\
\end{aligned} \right.$
$J=\left. \left( {{t}^{3}}+1 \right)\ln \left( t+1 \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\left( {{t}^{2}}-t+1 \right)dt}$ $=\left. \left( {{t}^{3}}+1 \right)\ln \left( t+1 \right) \right|_{0}^{1}-\left. \left( \dfrac{{{t}^{3}}}{3}-\dfrac{{{t}^{2}}}{2}+t \right) \right|_{0}^{1}=2\ln 2-\dfrac{5}{6}$.
Vậy $\int\limits_{\dfrac{1}{e}}^{e}{\dfrac{f\left( \ln x \right)}{x}dx}=2\sqrt{3}+2\ln 2-\dfrac{31}{6}$.
$\Rightarrow $ $a+b+6c=2+2-31=-27$.
Đáp án C.